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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. |
Salvarsan, an organic arsemic derivative, used in the treatment ofA. SyphilisB. TyphoidC. MeningitisD. Dysentry |
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Answer» Correct Answer - A Fact |
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| 2. |
79. Which one is the correct order of acidity(1) \( CH \equiv CH > CH _{3}- C \equiv CH > CH _{2}= CH _{2}> CH _{3}- CH _{3} \)(2) \( CH \equiv CH > CH _{2}= CH _{2}> CH _{3}- C \equiv CH > CH 3- CH \)(3) \( CH _{3}- CH _{3}> CH _{2}= CH _{2}> CH _{3}- C \equiv CH > CH \equiv CH \)(4) \( CH _{2}= CH _{2}> CH _{3}- CH = CH _{2}> CH _{3}- C \equiv CH > CH \equiv CH \) |
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Answer» correct answer is 1: Since terminal carbon of alkyne group is acidic because is has higher electronegativity then other and in CH3–C=_CH CH3 is showing +I effect which will un stabilise the - charge on C which will come after H is gone Order of Electronegativity:- alkyne > alkene > alkane |
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| 3. |
By how much will the potential of half-cell `Cu^(2+)|Cu` change if the solution is diluted to 100 times at 298K?A. Increases by 59 mVB. Decreases by 59 mVC. Increases by 29.5 mVD. Decreases by 29.5 mV |
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Answer» Correct Answer - B `Cu^(+2) (aq) + 2e^(-) to Cu(s)` `E_(RP) = E_(RP)^(@) -(0.0591)/(2) log ""(1)/([Cu^(+2)])` |
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| 4. |
For every `10^(@)` C increase in temperature the rate becomes twice so the temperature increase from `10^(@)`C then rate becomes …. Times :-A. 400B. 512C. 112D. 614 |
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Answer» Correct Answer - B `K_(2)=K_(1)(m)^(DeltaT//10)=K_(1) (2)^(90//10)=K_(1)xx(2)^(9)=512K_(1)` |
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| 5. |
Kolbe's reaction |
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Answer» Kolbe reaction, also known as Kolbe Schmitt Reaction is a type of addition reaction named after Hermann Kolbe and Rudolf Schmitt. When phenol is treated with sodium hydroxide, phenoxide ion is generated. The phenoxide ion generated is more reactive than phenol towards electrophilic aromatic substitution reaction. Hence, it undergoes electrophilic substitution reaction with carbon dioxide, which is a weak electrophile. Ortho-hydroxybenzoic acid (salicylic acid) is formed as the primary product. This reaction is popularly known as Kolbe’s reaction. |
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| 6. |
The rate contant for zero order reaction is `3xx10^(-2) mol^(-1)L.sec^(-1)`. After 25 sec. If the concentraition of reactant is 0.5 M them initial concentraition of the reactant is :-A. 1 MB. 1.25 MC. 0.5 MD. 0.75 M |
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Answer» Correct Answer - B `A_(0)-A=Kt` `A_(0)-0.5=3xx10^(-2)xx25` `A_(0)0.5+0.75=1.25M` |
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| 7. |
The plot of In 1/T is linear with slope of :A. `-(E_(a))/( R)`B. `+(E_(a))/( R)`C. `-(Ea)/(2.303R)`D. `(E_(a))/(2.303R)` |
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Answer» Correct Answer - A ln K ln `A-(Ea)/(RT)` ln K = ln `A-(Ea)/(R )((1)/(T))` slope `=(-Ea)/( R)` |
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| 8. |
In a reaction `N_(2(g))+3H_(2(g))to2NH_(3(g))` rate of appearance of `NH_(3)` is `2.5xx10^(-4)molL^(-1)sec^(-1)` then rate of reaction and rate of disappearance of `H_(2)` respectively is :-A. `3.75xx10^(-4),1.25xx10^(-4)`B. `1.25xx10^(-4),2.5xx10^(-4)`C. `1.25xx10^(-4),3.75xx10^(-4)`D. `5xx10^(-4),3.75xx10^(-4)` |
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Answer» Correct Answer - C `ROR=(-d[N_(2)])/(dt)=(-d[H_(2)])/(3dt)=(d[NH_(3)])/(2dt)` `ROR=(d[NH_(3)])/(2dt)` `ROR=(2.5xx154)/(2)=1.25xx10^(-4)` `(-d[H_(2)])/(dt)=(3)/(2)(d[NH_(3)])/(dt)=(3)/(2)xx2.5xx10^(-4)=3.75xx10^(-4)` |
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| 9. |
Give one example of a reaction where order and molecularity are equal ? |
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Answer» 2HI => H2 + I2 (Order = Molecularity = 2) |
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| 10. |
Three-fourth of a reaction is completed in 32 minutes. What is the half life period of this reaction ? |
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Answer» The answer is 16 minutes. |
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| 11. |
What is meant by an elementary reaction ? |
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Answer» A reaction which takes place in one step is called an elementary reaction. For |
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| 12. |
For a reaction R =>P, the rate becomes 2 times when the concentration of thereactant A is increased 4 times. What is the order of reaction ? |
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Answer» r = k(a)n => 2r = k(4a)n => 2 = 4n => n = 0.5 |
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| 13. |
What is meant by an elementary reaction ? |
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Answer» A reaction which takes place in one step is called an elementary reaction. For example : H2 + I2 => 2HI. |
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| 14. |
Give one example of a reaction where order and molecularity are equal ? |
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Answer» 2HI => H2 +l2 (Order =molecularity =2) |
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| 15. |
The rate of disappearence of `SO_(2)` in the reaction `2SO_(2)+O_(2)to2SO_(3)` is `1.28xx10^(-3)` g `litre^(-1) sec^(-1)` Then the rate of formation of `SO_(3)` in g `litre^(-1)` `sec^(-1)` isA. `0.64xx10^(3)`B. `0.80xx10^(3)`C. `1.28xx10^(-1)`D. `1.60xx10^(-3)` |
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Answer» Correct Answer - 4 `=-(d[SO_(2)])/(dt)=1.28xx10^(-3)gL^(-1)s^(-1)` `=(1.28xx10^(-3))/64Ms^(-1)` `-(d[SO_(2)])/(dt)=(d[SO_(3)])/(dt)` `(1.28xx10^(-3))/64ML^(-1)s^(-1)` `(1.28xx10^(-3))/64xx80gL^(-1)s^(-1)` `=1.6xx10^(-3)gL^(-1)s^(-1)` |
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| 16. |
For a reaction R=> P, the rate becomes 2 times when the concentration of the reactant A is increased 4 times. What is the order of reaction ? |
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Answer» r = k(a)n => 2r = k(4a)n =>2 = 4n => n = 0.5 |
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| 17. |
Rate of formation `SO_(3)` in this reaction is `1.6xx10^(-3)kg//min` `2SO_(2)+O_(2)to2SO_(3)` then rate at which ` SO_(2)` reacts is :-A. `1.6xx10^(-3) kg//mi n^(-1)`B. `8xx10^(-4) kg//mi n^(-1)`C. `3.2xx10^(-3)kg mi n^(-1)`D. `1.28xx10^(-3)kg mi n^(-1)` |
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Answer» Correct Answer - D `(dSO_(3))/(dt)=1.6xx10^(-3) Kg//min` `=(1.6xx10^(-3)1000)/(80)mol//min` `(dSO_(2))/(2dt)-(-dO_(2))/(dt)=(dSO_(3))/(2dt)` `(-dSO_(2))/(dt)=(1.6)/(80)mol//min` `(1.6)/(80)xx64xx10^(-3)kg//mol` `=1.28xx10^(-3)mol//min` |
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| 18. |
If concentration of reactant is increased by 2 times then K becomes :-A. `1n(K)/(2)`B. `(K)/(2)`C. 2KD. K |
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Answer» Correct Answer - D Rate constant is independent of concentration of Reactant` |
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| 19. |
Calculate the order of the reaction in A and B :- `{:(A,,,B,,,Rate),(("mol"//1),,,("mol"//1),,,),(0.05,,,0.05,,,1.2xx10^(-3)),(0.10,,,0.05,,,2.4xx10^(-3)),(0.05,,,0.10,,,1.2xx10^(-3)):}`A. 1 and 0B. 1 and 1C. 0 and 1D. None of these |
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Answer» Correct Answer - A `(Exp(1))/(Exp(2))(1.2xx10^(-3))/(2.4xx10^(-3))=(K(0.05)^(x)(0.05)^(y))/(K(0.10)^(x)(0.05)^(y))` `(1)/(2)=((1)/(2))^(x)(X=1)` `(Exp(1))/(Exp(3))(1.2xx10^(-3))/(1.2xx10^(-3))=(K(0.05)^(x)(0.05)^(y))/(K(0.05)^(X)(0.05)^(Y))` `1=((1)/(2))y" "y=0` |
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| 20. |
How many moles of alkene are required with one mole of diboarane in the formation of alcohol from hydroboration-oxidation. |
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Answer» Diborane (B2H6) react with an Alkene to give trialkyl borane as an addition product which further oxidized by H2O2 to alcohol. As we know, 1 mole B2H6 (Diborane) → 2BH3 (2 mole) ∴ 1 mole diborane can oxidized 6 mole of alkene into alcohol. |
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| 21. |
The half life for the first order reaction `AtoB+C` is 24 hrs. starting with 10g of A how many grams of A will remain after 96 hours :-A. 1.25 gB. 0.63 gC. 1.77 gD. 0.5 g |
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Answer» Correct Answer - B `10gramoverset(24hrs)to5goverset(24hrs)to2.5goverset(24hrs)to1.25goverset(24hrs)to0.626gram` |
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| 22. |
A solution containing 6 g of a solute is dissolved in 25mL of water gave an osmotic pressure of 4.5 atm at 27°C. Calculate the boiling point of the solution. Kb = 0.52Kkg/mol. |
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Answer» Let say molar mass of solute is M g/mol. Then \(\pi = \frac{\frac6M}{25} \times 1000\times R\times T\) \(\pi = 4.5 = \frac6{M\times25} \times 1000\times0.082 \times300\) \(M = \frac6{4.5} \times 40 \times 0.082 \times 300\) \(m = 1312 g/ mol\) ∴ Molality of solution \(= \frac{\frac6{1312}}{25} \times 1000 = 0.183 m\) ∴ \(\triangle T_b = K_bm\) \(T - 100 = 0.52 \times 0.183\) \(T = 100.095° C\) Hence, boiling point of solution becomes 100.095°C. |
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| 23. |
a) \( 2 \cos ^{2} \theta-1 \)b) 1-2 \( \sin ^{2} \theta \)c) \( \cos ^{2} \theta-\sin ^{2} \theta \)d) \( \cos ^{2} \theta+\sin ^{2} \theta \)6. \( \cos ^{2} \theta= \)a) \( \frac{1+\cos 2 \theta}{2} \)b) \( \frac{1-\cos 2 \theta}{2} \)c) \( 1-\sin ^{2} \theta \)d) \( \cos (2 \theta) \) |
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Answer» sin2x= 2sinx . cosx |
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| 24. |
Give balanced equation and name of products when CH3COOH is treated with PCl5 ? |
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Answer» CH3COOH + PCl5 => CH3COCl + POCl3 + HCl |
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| 25. |
Write the name of ester which gives the ¯avor of pineapple. |
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Answer» Ethyl butanoate. |
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| 26. |
Write the name of the cell which is generally used in hearing aids. Write the reactions taking place at the anode and cathode of this cell. |
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Answer» A mercury battery, used for hearing aids and electric watches, delivers a constant voltage of 1.35 V for long periods. The half-reactions are Anode: Zn (Hg) + 2 OH ̶ → ZnO (s) + H2O + 2e ̶ Cathode: HgO + H2O + 2e ̶ → Hg (l) + 2 OH ̶ |
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| 27. |
Calculate the density of Nacl if the edge length of Nacl unit cell is 564pm |
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Answer» answer : 2.17 g/cm^3 |
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| 28. |
What product is obtained when ethyl benzene is oxidized with alkaline KMnO4 ? |
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Answer» Benzoic acid (C6H5COOH) is obtained. |
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| 29. |
The number of moles of mno4- reduced to mn+2 by the addition of 7.5 mole electron in mno4- is A)2.5 B)5 C)1.5 D) 7.5 |
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Answer» OPTION (C) IS CORRECT Ionic redox reaction mentioned here is represented by the following balanced equation |
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| 30. |
Name the raw materials added before smelting, in the extraction of iron ore? |
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Answer» Lime Stone is added to Blast Furnace before smelting. Limestone and coke are the raw materials added before smelting in blast furnace in the extraction of iron ore. |
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| 31. |
HCOOH , C6H5COOH , CH3COOH |
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Answer» Higher the stability of conjugate base higher is the acidity. So, Acidic Strength order is HCOOH>C6H5COOH >CH3COOH |
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| 32. |
Elevation of boiling point is inversely proportional to :(A) molal elevation constant (Kb)(B) molality (m)(C) molar mass of solute (M)(D) weight of solute (W) |
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Answer» Option : (C) molar mass of solute (M) ΔTb = kbm or, ΔTb = kb\(\frac{W_1}{M\times W_2}\), Where Kb = molal elevation constant, m = molality or, ΔTb = kb\(\frac{W_1}{M\times W_2}\), Where W1 = mass of solute,W2 = mass of solvent, M = molar mass of solute. ∴ ΔTb ∝ \(\frac{1}{M}\) Therefore, Elevation of boiling point is inversely proportional to molar of solute (M). |
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| 33. |
Chlorine reacts with cold and dilute NaOH to give :(A) Nacl and NaClO3(B) Nacl and NaClO(C) Nacl and NaClO4(D) Nacl and NaClO3 |
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Answer» Option : (B) Nacl and NaClO When cold and dilute NaOH react with Cl2 gas. yields. Nacl,Naclo and H2O 2 NaOH + Cl2 → Nacl + NaOCl + H2O Sodium hypochloride |
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| 34. |
In a hydrocarbon the mass ratio of hydrogen to carbon is `1:3`. The empirical formula of the hydrocarbon isA. (a)`CH_(2)`B. (b)`C_(2)H`C. (c )`CH_(3)`D. (d)`CH_(4)` |
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Answer» Correct Answer - D In `CH_(4)` carbon is `12` and hydrogen is `4xx1` |
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| 35. |
The position vector of a aprticle is given as `vecr=(t^2-4t+6)hati+(t^2)hatj`. The time after which the velocity vector and acceleration vector becomes perpendicular to each other is equal toA. `1 sec`B. `2 sec`C. `1.5 sec`D. not possible |
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Answer» Correct Answer - A `(A)` `vec(r)=(t^(2)-4t+6)hat(i)=t^(2)hat(j),vec(v)=(dvec(r))/(dt)=(2t-4)hat(i)+2that(j), vec(a)=(dvec(v))/(dt)=2hat(i)+2hat(j)` if `vec(a)` and `vec(v)` are perpendicular `vec(a).vec(v)=0` `(2hat(i)+2hat(j)).((2t-4)hat(i)+2that(j))=0` `8t-8=0` `t=1sec.` `Ans. t=1 sec`. |
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| 36. |
The block of mass `m` initially at `x=0` is acted upon by a horizontal force at any position `x` is given as `F =a-bx^(2)(` where `a gt mu mg)`, as shown in the figure. The co`-` efficient of friction between the surface of contact is `mu`. The net work done on the block is zero, if the block travels a distance of `......................` |
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Answer» Correct Answer - `x=[3(a-mu mg)//b]^(1//2)` `W_(f)+W_(F)=-mu mg x+underset(0)overset(x)int (a-bx^(2))dx` `=(-mu mg +a)x-(bx^(3))/(3)` `rArr x=sqrt((3(a-mu mg))/(b))` |
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| 37. |
A body starts from rest and moves with constant acceleration. The ratio of distance covered by the body in `nth` second to that covered in `n` second is.A. `(2)/(n^(2))-(1)/(n)`B. `(2)/(n^(2))+(1)/(n)`C. `(2)/(n)-(1)/(n^(2))`D. `(1)/(n)+(1)/(n_(2))` |
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Answer» Correct Answer - C `(S_(N))/(S)=((1)/(2)a(2n-1))/((1)/(2)an^(2))=(2n)/(n^(2))-(1)/(n)-(1)/(n^(2))`. |
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| 38. |
A body starts from rest and moves with constant acceleration. The ratio of distance covered by the body in `nth` second to that covered in `n` second is.A. `(2)/(n^(2))-(1)/(n)`B. `(2)/(n^(2))+(1)/(n)`C. `(2)/(n)-(1)/(n^(2))`D. `(2)/(n)+(1)/(n^(2))` |
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Answer» Correct Answer - C `(S_(N))/(S)=((1)/(2)a(2n-1))/((1)/(2)an^(2))=(2n)/(n^(2))-(1)/(n^(2)), (2)/(n)-(1)/(n^(2))` |
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| 39. |
Two bikes `A and B` start from a point. A moves with uniform speed `40 m//s and B` starts from rest with uniform acceleration `2 m//s^2`. If `B` starts at `t = 10` and `A` starts from the same point at `t = 10 s`, then the time during the journey in which `A` was ahead of `B` is :A. `20s`B. `8s`C. `10s`D. `A` is never ahead of `B` |
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Answer» Correct Answer - D `A` will be ahead of `B` when `X_(A)gtX_(B)` `40(t-10)gt(0)t+(1)/(2)(2)t^(2)` as `A` is `10sec. `late than `B`. `rArr t^(2)-40t+4 lt0` `rArr (t-20)^(2) lt0` Which is not possible . So `A` will never to ahead at `B`. |
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| 40. |
For what MANOVA is used for?(a) Detection of biological samples(b) Detection of the best fit curve(c) Comparison between monovariate samples(d) Comparison between multivariate samples |
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Answer» The correct option is (d) Comparison between multivariate samples To explain I would say: In statistics, multivariate analysis of variance (MANOVA) is a procedure for comparing multivariate sample means. As a multivariate procedure, it is used when there are two or more dependent variables and is typically followed by significance tests involving individual dependent variables separately. |
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| 41. |
For the pelleting of the microsomal fraction from a liver homogenate, an ultracentrifuge is operated at a speed of 40000 rpm. What is the angular velocity, ω, in radians per second?(a) 3888.8 rad s^-1(b) 5680.8 rad s^-1(c) 4188.8 rad s^-1(d) 4288.8 rad s^-1 |
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Answer» Correct answer is (c) 4188.8 rad s^-1 The best explanation: The angular velocity, ω, may be calculated using the equation: ω = (2 π rev min-1)/60 ω = 2 × 3.1416 × 40000/60 rad s^-1 = 4188.8 rad s^-1. |
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| 42. |
A ball is thrown onto a smooth floor with speed `u` at angle `theta = 45^(@)`. If it rebounds with a speed `v` at the same angle `phi = 45^(@)`. Then the coefficient of restitution is A. `0.5`B. `0.6`C. `0.1`D. `1` |
| Answer» Correct Answer - D | |
| 43. |
If `tan 0 gt mu` then minimum force needed to stay block at rest is A. `mg sin theta - mu mg cos theta`B. `(mg sin theta - mu mg cos theta)/(sqrt(1+mu^(2)))`C. `mg sin theta+mu mg cos theta`D. Zero |
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Answer» Correct Answer - B If `tan theta gt mu` …………. `F cos phi+muN = mg sin theta` `N = mg cos theta+F sin phi` `F = (mg sin theta - mu mg cos theta)/(mu sin phi+cos phi)` |
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| 44. |
An elastic string has a force constant k and mass m. the string hangs vertically, and a block of an unknown mass is attached to its bottom end. It is known that the mass of the blocks is much greater than that of the string. The hanging block stretches the string to twice its relaxed length. how long (t) would it t take for a low-amplitude transwerse pulse to travel the length of the string stretched by the hanging block ? m=1 kg, `k=1/2 N//m`. |
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Answer» Correct Answer - 2 `T=Mg = k Deltal =kxxl` `t=(2l)/v =(2l)/(sqrt(T/(mu))) =21/(sqrt((kl)/(m//2l)))` `(2l)/(sqrt(2kl^(2)//m))=sqrt((2m)/k) =2 sec` |
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| 45. |
A uniform disc of mass `m` & radius `R` is pivoted at its centre `O` with its plane vertical as shown in figure `A` circular portion of disc of radius `(R)/(2)` is removed from it. Then choose the correct option(s) A. Time period of small oscillations of remaining portion about `O` is `pisqrt((13R)/(g))`B. Time period of small oscillations of remaining portion about `O` is `2pisqrt((13R)/(g))`C. The centre of mass of the remaining disc is at a distance of `(R)/(6)` from `O`.D. The centre of mass of the remaining disc is at a distance of `(R)/(8)` from `O`. |
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Answer» Correct Answer - A::C `T = 2pisqrt((I)/(mgy_(cm)))` `t = (MR^(2))/(2) - ((1)/(2)(M)/(4)((R)/(2))^(2) + (M)/(4)((R)/(2))^(2)) = (13MR^(2))/(32)` `rArr y_(cm) = (R)/(6)` `m = (3M)/(4)c` `T = 2pi , sqrt((1)/(4)(R)/(g))` |
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| 46. |
A uniform disc of mass `m` & radius `R` is pivoted at its centre `O` with its plane vertical as shown in figure A circular portion of disc of radius `(R )/(2)` is removed from it. The time period of small oscillations of remaining portion about `O` is - A. `3pi sqrt((R )/(g))`B. `pi sqrt((13R)/(g))`C. `2pi sqrt((39R)/(16g))`D. `2pisqrt((7R)/(6g))` |
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Answer» Correct Answer - B A uniform ………… `T = 2pi sqrt((I)/(mgy_(cm)))` `I = (MR^(2))/(2) - ((1)/(2)(M)/(4)((R )/(2))^(2) + (M)/(4)((R )/(2))^(2))= (13MR^(2))/(32)` `y_(cm) = (R )/(6)` `T = 2pi sqrt((13)/(4) (R )/(g))` |
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| 47. |
The total energy of the body excuting `S.H.M.` is `E` . Then the kinetic energy when the displacement is half of the amplitude, isA. `(E )/(2)`B. `(E )/(4)`C. `(3E)/(4)`D. `(sqrt(3))/(4)E` |
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Answer» Correct Answer - C The total ………….. kinetic energy `=(1)/(2)m omega^(2)(a^(2)-x^(2))` kinetic energy `= (1)/(2) ((3)/(4)m omega^(2)a^(2)) = (3E)/(4)` |
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| 48. |
A spherical shell of mass `M` and radius `R` filed completely with a liquid of same mass and set to rotate about a vertical axis through its centre has a moment of inertia `I_(1)` about the axis. The bottom. If moment of inertia of the system is `I_(2)` when the shell is half filled and `I_(3)` is the moment of inertia when entire water drained off then :A. `(I_(1))/(I_(2)) ~~ 1.5`B. `(I_(1))/(I_(2)) ~~0.67`C. `(I_(1))/(I_(3)) ~~ 1.6`D. `(I_(2))/(I_(3)) ~~ 1.4` |
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Answer» Correct Answer - C A spherical shell …………. `(C ) I_(1) = (2)/(3)MR^(2)+(2)/(5)MR^(2)= (16)/(15)MR^(2)` `I_(2) = (2)/(3) MR^(2)+(2)/(5)(M)/(2)R^(2)=(13)/(15)MR^(2)` `I_(3)=(2)/(3)MR^(2)` , `(I_(1))/(I_(3)) = 1.6` |
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| 49. |
A uniform disc of mass `M` and radius `R` is liffted using a string as shown in the figure. Then choose incorrect option(s), A. its linear acceleration is g upwardB. its linear acceleration is g downwardC. its angular acceleration is `(2g)/(R )` .D. its rate of change of angular momentum is `MgR` . |
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Answer» Correct Answer - B A uniform disc …………….. `(3Mg)/(2)+(Mg)/(2)-Mg=Ma` `a = g uarr` `tau = I alpha` `(3Mg)/(2)R -(Mg)/(2)R = (MR^(2))/(2).alpha` `2Mg alpha = (2g)/(R )` |
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| 50. |
A uniform disc of mass `M` and radius `R` is liffted using a string as shown in the figure. Then choose incorrect option(s), A. its linear acceleration is g upwardB. its linear acceleration is g downwardC. its angular acceleration is `(2g)/(R)`D. its rate of change of angular momentum is `MgR` |
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Answer» Correct Answer - A::C::D `(3Mg)/(2)+(mg)/(2)-Mg=Ma` `a=g uarr tau = I prop` `(3Mg)/(2)R-(Mg)/(2)R=(MR^(2))/(2) prop` `2Mg prop = (2g)/(R)rArr (dL)/(dt)=tau = MgR`. |
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