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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 25851. |
Among the following the most stable compound is:(1) Cis-1, 2-cyclohexanediol (2) trans-1, 2-cyclohexanediol (3) cis-1, 3-cyclohexanediol (4) trans-1, 3-cyclohexanediol |
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Answer» The most stable compound is trans-1, 3-cyclohexanediol |
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| 25852. |
A horizonal piston-cylinder arrangement is placed in a constant temperature bath.The piston slides in the cylinder with negligible friction, and an external force holds it in place against an initial gas pressure of 14 bar.The initial gas volume is `0.03 m^3` (a)The external force on the piston is reduced gradually, allowing the gas to expand until its volume doubles.Calculate the work done by the gas in moving the external force. (b)How much work would be done if the same expansion is carried out by removing a part of the external force suddenly.Also calculate % efficiency of this process as compared with the reversible process (%of reversible work). Take `ln2=0.7` Report your answer as (% efficiency `xx 0.7`). |
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Answer» Correct Answer - 50 (a)W=-nRT In `V_2/V_1` `W=-P_1V_1` In `V_2/V_1=-14xx0.03"In"0.06/0.03"bar"m^3= -14xx0.7xx0.03= -0.294 "bar" m^3` (b)`P_1V_1=P_2V_2` `:. P_2=(P_1V_1)/V_2=(14xx0.03)/(0.06)=7` bar `:. W=-P_(ext)(V_2-V_1)=-7(0.06-0.03)=-7xx0.03= -0.21 "bar"m^3` Efficiency `=0.21/0.294=71.43%` |
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| 25853. |
Only `N_2,CO and CO_2` gases remain after 0.72 gm of carbon is treated with one litre of air at `27^@C and 4.92` atm pressure.Assume air composition `O_2=20%,N_2=79%,and CO_2=1%` (by volume).The heat evolved (in Kcal) under constant pressure is : Given : `C+OtoCO_2, DeltaH=-100` Kcal/mol `C+1/2O_2toCO, DeltaH=-25` Kcal/mol |
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Answer» Correct Answer - Heat evolved = 3 kcal `n_(air)=(4.92xx1)/24.6=0.2` mol `n_(O_2)=0.2xx0.2=0.04 mol, n_c=0.72/12` `{:(,C" "+,1/2O_2,toCO),("initial mole",0.06,0.04," " 0),("final mole",0,0.01," "0.06):}` `{:(,CO" "+,1/2O_2,toCO_2),("initial mole",0.06,0.01," " 0),("final mole",0.04,0," "0.02):}` Heat evolved =`0.04xx25+0.02xx100=1+2=3 kcal` |
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| 25854. |
Light of the eye lens increases when eye musles...1.are relaxed and lens becomes thinner.2.contract and lens becomes thicker.3.are relaxed and lens becomes thicker.4.contract and lens becomes thinner. |
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Answer» 4.contract and lens becomes thinner. |
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| 25855. |
The turning back of light into the same medium after incident on a boundary separating two media is called(a) reflection of light (b) refraction of light(c) dispersion of light (d) interference of light |
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Answer» (a) reflection of light |
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| 25856. |
Electromagnetic radiation belonging to _________ region of spectrum is called light.(a) 100 nm to 400 nm (b) 400 nm to 750 nm(c) 750 nm to 10 nm (d) 1000 nm to 1400 nm |
| Answer» (b) 400 nm to 750 nm | |
| 25857. |
Give the uses of carbon dioxide. |
Answer»
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| 25858. |
Predict the conditions under which(а) Aluminium might be expected to reduce magnesia. (b) Magnesium could reduce alumina. |
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Answer» The conditions under which: (a) Ellingham diagram is used to predict thermodynamic feasibility of reduction of oxides of one metal by another metal. Any metal can reduce the oxides of other metals that are located above it in the Ellingham diagram. In the Ellingham diagram, for the formation of magnesia (magnesium oxide) occupy lower position than aluminium oxide. Therefore aluminium cannot be used to reduce the oxides of magnesium (magnesia). Above 1623K, A1 can reduce MgO to Mg, so that ArG° becomes negative and the process becomes thermodynamically feasible. 3MgO + 2Al + 1623 k ⟶ Al2O3 + 3 Mg (b) 1. \(\frac{4}{3}\)+ Al + O2 ⟶ \(\frac{2}{3}\)Al2O3 2. 2Mg + O2 ⟶ 2MgO At the point of intersection of the Al2O3 and MgO curves in Ellingham diagram. ∆G° becomes zero for the reaction. \(\frac{2}{3}\)+ Al2O3 + 2Mg ⟶ 2 MgO + \(\frac{4}{3}\)Al |
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| 25859. |
Chlorophyll contains 4.8% Mg by mass. Calculate number of Mg atom in 2000 g chlorophyll (atomic weight Mg = 24) :-A. `N_A`B. `2N_A`C. `4 N_A`D. `6 N_A` |
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Answer» Correct Answer - C 100 g chlorophyll `to` 4.8 g Mg 2000 g Chlorophyll `to 4.8/100xx2000` = 96 g Mg Number of Mg atoms = Moles x `N_A` `="Mass"/"Molar mass" xx N_A` `=96/24xxN_A` `=4N_A` |
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| 25860. |
A will be more stable than B if X is :A. `-CH_3`B. `-C_2H_5`C. `-F`D. `-Br` |
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Answer» Correct Answer - C Due to intramolecular H-bond (A) is more stable than B |
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| 25861. |
Name three important crystalline forms of silica. |
| Answer» Quartz,tridymite and crystobalite. | |
| 25862. |
Given: (i) `Cu^(2+)+2e^(-) rarr Cu, E^(@) = 0.337 V` (ii) `Cu^(2+)+e^(-) rarr Cu^(+), E^(@) = 0.153 V` Electrode potential, `E^(@)` for the reaction, `Cu^(+)+e^(-) rarr Cu`, will beA. `0.90 V`B. `0.30 V`C. `0.38 V`D. `0.52 V` |
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Answer» Correct Answer - D `Cu^(+2)+2e^(-) to Cu " " DeltaG_(1)^(@)` `(Cu^(+2) +e^(-) to Cu^(+) " "DeltaG_(2)^(@))/(Cu^(+) +e^(-) to Cu " " DeltaG_(3)^(@))` `DeltaG_(3)^(@) =DeltaG_(1)^(@) -DeltaG_(2)^(@)` `-1xxF xx E =- 2F xx 0.337 -(-1xxF 0.153)` `rArr E= 0.521 V` |
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| 25863. |
Give two reasons, why the number of elements in the first period is only `2`? |
| Answer» It is because the `1 st` energy level can have only `1s` orbital which can have two electrons. When `n=1`,`l=0`. | |
| 25864. |
Why is `CO` combustible and `CO_(2)` non-combustable? |
| Answer» `CO` is combustible because it has a great affnity for oxygen and burns in `O_(2)` to form `CO_(2)`. This reaction is exothermic. `CO_(2)` has no affinity for oxygen and so it is non-combustible. The non-combustible nature of `CO_(2)` makes it a good fire extinguisher. | |
| 25865. |
`HNO_(3)` has no action of aluminium, whether it is dilute or concentrated? |
| Answer» Because the metal is rendered passive. | |
| 25866. |
Name the following boric acids: a.`H_(3)BO_(3)` or `B_(2)O_(3).3H_(2)O` b.`HBO_(2)` or `B_(2)O_(3).H_(2)O` c. `H_(2)B_(4)O_(7)` or `2B_(2)O_(3).H_(2)O` d. `H_(6)B_(4)O_(9)` or `2B_(2)O_(3).3H_(2)O` |
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Answer» a. Boric or orthoboric acid b. Metaboric acid c. Tetraboric acid d. Pyro-boric acid |
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| 25867. |
On the basis of their electronic configuration, explain why alkali metals are highly reactive? |
| Answer» Alkali metlas have general electronic configuration `ns^(1)`. They can lose `1` electron to acquire stable electronic configuration. They have large atomic size, therefore, they can lose electron easily and they are most reactive. | |
| 25868. |
The EAN of metal atoms in `[Fe(XO)_2(NO^+)_2]` and `Co_2 (CO)_8` respectively are :A. 34, 35B. 34, 36C. 36, 36D. 36, 35 |
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Answer» Correct Answer - C Oxidation state of iron in `overset("-2 " " 0 " " +)([Fe(CO)_2(NO)_2])`=x+2(0)+2=0,x= -2 So EAN =28+8=36 Oxidation state of cobalt in `Co_2(CO)_8`=2x+8(-0)= or x=0 So EAN = 27+1+8 =36 |
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| 25869. |
What does the retention factor, k’, describe?(a) The velocity from the stationary phase(b) The velocity of the mobile phase(c) The distribution of an analyte between the stationary and the mobile phase(d) The migration rate of an analyte through a column |
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Answer» Correct answer is (d) The migration rate of an analyte through a column Best explanation: k’ (Capacity factor) in the chromatography is to provide a calculation or formula which defines how much interaction the solute has with the stationary phase material. And it is based on the formula given below: Where, T (R) = Retention time of the peak in minutes T (0) = Retention time of an unretained peak k’ value should be >1. |
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| 25870. |
On boiling with concentrated HBr phenyl ethyl ether will give.A. Phenol and ethyl bromideB. Bormobenzene and ethanolC. Phenol and ethaneD. Bromobenzene and ethane. |
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Answer» Correct Answer - A `C_6H_5-O-CH+`conc. `Hbroverset(Delta)toC_6H_5-OH+CH_2Br` |
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| 25871. |
One pink solid (X) has the formula `CoCl_3. 5NH_3. H_2O`.A solution of this salt (X) in water is also pink and precipitates 3 moles AgCl on treatment with silver nitrate solution..When the pink solid (X) is heated, it loses 1 mole `H_2O` to yield a purple solid (Y) with the same molar ratio of `NH_3:Cl:Co`.The purple solid releases two of its chlorides rapidly on dissolution and after treatment with `AgNO_3` solution.The coordination number of cobalt in complexes (X) and (Y) is six. The IUPAC name of pink (X) and purple (Y) complexes respectively are :A. Tetraamminechloridoaquacobalt(III) chloride and pentaamminechloridocobalt(III) chlorideB. Pentaammin echloridocarbalt(III) chloride and pentaammineaquacobalt (III) chlorideC. Pentaammineaquacobalt(III) chloride and pentaamminechloridocobalt(III)chlorideD. Pentaammineaquacobaltate(III) chloride and pentaamminechloridocobaltate (III) chloride |
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Answer» Correct Answer - C `underset(1:5:1)([Co(NH_3)_5H_2O])Cl_3("pink")overset(aq.)hArr[Co(NH_3)_5H_2O]^(3+)+3Cl^(-)` `3Ag^(+)+3 Cl^(-)to3AgCldarr`(white) The following complex is obtained on heating and losing 1 mole of `H_2O` `underset(1:5:1)([Co(NH_3)_5Cl]Cl_2)("purple")overset(aq.)hArr[Co(NH_3)_5Cl]^(2+)+2Cl^(-)` `2Ag^(+)+2Cl^(-)to2AgCldarr`(white) So, pink (X) is `[Co(NH_3)_5(OH_2)]Cl_3` and IUPAC name would be pentaammineaquacobalt(III) chloride and purple (Y) is `[Co(NH_3)_5Cl]Cl_2` and IUPAC name would be pentaamminechloridocobalt(III)chloride. |
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| 25872. |
Rotary drum tank reactor comes under which type of bioreactors?(a) Mechanically agitated bioreactors(b) Air driven bioreactors/Pneumatically agitated bioreactors(c) Non-agitated bioreactors(d) Wave bioreactors |
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Answer» The correct answer is (a) Mechanically agitated bioreactors To explain I would say: Rotary drum tank reactor comes under Mechanically agitated bioreactors because it contains Stirrer paddle which provides efficient mixing and also causes production on a large scale. |
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| 25873. |
Which of the following is not characteristic of chemisorption?(a) It is irreversible(b) It is specific(c) It is multilayer phenomenon(d) Heat of adsorption is about 400kj |
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Answer» Correct choice is (c) It is multilayer phenomenon To explain I would say: Chemisorption involves formation of chemical bonds between adsorbate and adsorbent molecules. Once the valency is satisfied, the adsorbent molecules can’t form bond with more adsorbate molecules. Thus only one layer is formed. |
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| 25874. |
What do you mean by the term “Sorption”?(a) Attachment(b) Detachment(c) Diffusion(d) Thermal Expansion |
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Answer» The correct answer is (a) Attachment The best I can explain: Sorption is a physical and chemical process by which one substance becomes attached to another. The reverse of sorption is desorption. |
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| 25875. |
Match the complexes listed in column-I with characteristic type of isomerism listed in column-II `{:("Column I","Column II"),((A)CoCl_3. 5NH_3. H_2O,(p)"In one of this isomeric forms, the metal satisfies its secondary valences by only neutral molecules."),((B)Co(NH_3)_5BrSO_4,(q)"The magnetic moment of one of the isomers form is 3.87 B.M."),(( C)CrCl_3. 6H_2O,(r)"Show ionisation isomerism"),((D)Cr(NO_2)_2. 6H_2O,(s)"Show hydrate isomerism"),(,(t)"Show linkage isomerism."):}` |
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Answer» Correct Answer - A-p,s ; B-r ; C-p,q,s ; D-p,s,t (A)`overset(III)([Co(NH_3)_5Cl])Cl_2. H_2O and overset(III)([Co(NH_3)_5H_2O])Cl_3`-hydrate isomerism. In later form all (i.e. six) secondary valences of metal are satisfied by neutral molecules. (B)`overset(III)([Co(NH_3)_5Br])SO_4 and overset(III)([Co(NH_3)_5SO_4])`Br-ionisation isomerism. (C )`[Cr(H_2O)_6]Cl_3, [Cr(H_2O)Cl]Cl_2. H_2O` and `[Cr(H_2O)_4Cl_2]Cl.2H_2O`-hydrate isomerism. In first form secondary valences (i.e. six) of the metal are satisfied by only neutral molecules. Cr(III) has 3 unpaired electrons whether the ligand is weak or strong field , `mu_(B.M.)=sqrt(3(3+2))`=3.87 B.M. (D)`[Cr(H_2O)_6]NO_2 and [Cr(H_2O)_5 (NO_2)]NO_2 H_2O`-hydrate isomerism `[Cr(H_2O)_5(NO_2)]NO_2. H_2O and [Cr(H_2O)_5(ONO)]NO_2. H_2O` -linkage isomerism. In first form secondary valencies (i.e. six) of the metal are satisfied by only neutral molecules. In the complex the Cr is in +1 oxidation state with electron configuration. `[Ar]^(18)3d^5 4s^0`.It contains five unpaired electrons |
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| 25876. |
If curves y2 = 4x and xy = c cut at right angles, then the value of c is :(a) 4√2(b) 8(c) 2√2(d) -4√2 |
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Answer» Option : (a) 4√2 |
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| 25877. |
Match the following column of precipitate/mass listed in Column I with the reagent (s) listed in Column II: |
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Answer» Correct Answer - A::B::C::D `Pb^(2+):Pb^(2+)tounderset("Yellow")(Pbl_(2))darr+2K^(+)` `Pb^(2+)+K_(2)CrO_(4)tounderset("Yellow")(PbCrO_(4))darr+2K^(+)` `Ag^(+): 2Ag^(+)+2NaOHto underset("Brown")(Ag_(2)O)darr+H_(2)O+2Na^(+)` `2Ag^(+)+H_(2)Stounderset("Black")(Ag_(2)Odarr+2H^(+)` `Hg_(2)^(2+):H_(2)^(2+)+2NaOHtounderset("Black")(Hg_(2)O)+H_(2)O+2Na^(+)` `Hg_(2)^(2+)+H_(2)Sto underset("Black")ubrace(Hg darr+HgSdarr)+2H^(+)` `Hg_(2)^(2+)+2Kl to Hg_(2)l_(2)+2K^(+)` `Hg_(2)l_(2)+2Kl to K_(2)[Hgl_(4)]+underset("Black")(Hg)darr` |
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| 25878. |
The inverse of the matrix X = \(\begin{bmatrix}2& 0 & 0 \\[0.3em]0 & 3 & 0 \\[0.3em]0 & 0 & 4\end{bmatrix}\) is :[(2,0,0),(0,3,0),(0,0,4)](a) \(24\begin{bmatrix}1/2& 0 & 0 \\[0.3em]0 & 1/3 & 0 \\[0.3em]0 & 0 & 1/4\end{bmatrix}\)(b) \(\frac{1}{24}\begin{bmatrix}1& 0 & 0 \\[0.3em]0 & 1 & 0 \\[0.3em]0 & 0 &1\end{bmatrix}\) (c) \(\frac{1}{24}\begin{bmatrix}2& 0 & 0 \\[0.3em]0 & 3 & 0 \\[0.3em]0 & 0 &4\end{bmatrix}\) (d) \(\begin{bmatrix}1/2& 0 & 0 \\[0.3em]0 & 1/3 & 0 \\[0.3em]0 & 0 & 1/4\end{bmatrix}\) |
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Answer» Option : (d) \(\begin{bmatrix}1/2& 0 & 0 \\[0.3em]0 & 1/3 & 0 \\[0.3em]0 & 0 & 1/4\end{bmatrix}\) |
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| 25879. |
Select the incorrect statement about (A) and ( C) compounds.A. Aqueous solution of (C ) produces white precipitate with dilute HCl (2M)B. Aqueous solution of (A ) gives black white precipitate with `H_2S` gas and dilute HClC. Aqueous solution of (C ) produces red precipitate with `K_2CrO_4` solutionD. Aqueous solution of (A ) develops deep blue colouration with excess `NH_3` solution |
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Answer» Correct Answer - C `Pb^(2+)+CrO_4^(2-)toPbCrO_4 darr` (yellow) |
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| 25880. |
A chemist opened a cupboard and found four bottles containing water solutions, each of which had lost its label.Bottles 1,2,3 contanied colourless solution, while bottle 4 contained a blue solution.The labels from the bottles were lying scattered on the floor of the cupboard.They were: copper (II) sulphate, Hydrochloric acid lead nitrate , Sodium carbonate By mixing samples of the contents of the bottles, in pairs , the chemist made the following observations : Bottle 1 +Bottle 2 `to` White precipitate is formed. Bottle 1 +Bottle 3 `to` White precipitate is formed. Bottle 1 +Bottle 4 `to` White precipitate is formed. Bottle 2 +Bottle 3 `to` Colourless and odourless gas is evolved. Bottle 2 +Bottle 4 `to` No visible reaction is observed. Bottle 3 +Bottle 4 `to` Blue precipitate is formed. With the help of the above observations answer the following questions. Which of the following statements is correct for salts contained in bottle 1 and 4 ?A. Bottle 4 gives white precipitate with excess of KI solutionB. Bottle 4 gives white precipitate with excess of `K_4 [Fe (CN)_6]` solutionC. Bottle 1 and 4 both gives precipitate with excess of NaOH solutionD. Bottle 1 gives white precipitate with concentrated HCl solution |
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Answer» Correct Answer - A (A)`2Cu^(2+)+4I^(-)toCu_2I_2 darr ("white")+I_2` (B)`2Cu^(2+)+[Fe(CN)_6]^(4-)toCu_2[Fe(CN)_6]darr` (chocolate brown) (C )`Pb^(2+)+4OH^(-)to[Pb(OH)_4]^(2-)`(colourless soluble complex) , `Cu^(2+)+2OH^(-)toCu(OH)_2darr`(blue) (D)`Pb^(2+)+4Cl^(-)to[PbCl_4]^(2-)darr` (colourless soluble complex) |
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| 25881. |
What do you understand by budding? Write with examples. |
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Answer» Budding: It is a process in which a daughter individual is formed from a small projection. the bud, arising from the parent body. It takes place in Yeast, Sycon, Hydra etc. Budding is two types: (a) Exogamous budding: In which an outgrowth or bud grows externally on the body surface which may or may not split away from the surface of body. e.g. sycon. (b) Endogamous budding: In which buds are formed within the parent's body. They are called gemmules. Gemmules consist of small groups of cells enclosed by a protective coat. During favourable condition, the mass of archaeocytes comes out through micropyle and later on forms new colony. |
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| 25882. |
State two factors by which the range of transmission of TV signal can be increased. |
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Answer» Range of transmission of TV signal can be increased by- (i) Amplitude or Frequency modulation. (ii) Increasing the height of antenna. |
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| 25883. |
State two factors by which the range of transmission of TV signal can be increased. |
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Answer» Two factors are as follow— 1. Range of transmission of TV signal can be increased by increasing the height of antenna. 2. By Amplitude modulation |
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| 25884. |
Restriction enzymes are known as - (A) Biological guns (B) Molecular scissors (C) Plasmid (D) Micro pipette |
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Answer» (B) Molecular scissors |
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| 25885. |
Water holding capacity is one of the qualities of - (A) Soil (B) Plants (C) Water (D) Animals |
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Answer» The Correct option is (A) Soil |
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| 25886. |
Sickle-cell anemia is related to which type of disease? (A) Sex linked disease (B) Autosomal linked disease (C) Deficiency disease (D) Metabolic disease |
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Answer» (B) Autosomal linked disease |
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| 25887. |
Lac operon represent- (A) Inducible gene system (B) Repressible gene system (C) Housekeeping gene system D) All of these |
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Answer» (B) Repressible gene system |
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| 25888. |
A substance which gives dark red flame and breaks down on heating to give oxygen and a brown gas is:-A. `Mg(NO_(3))_(2)`B. `NaNO_(3)`C. `Ba(NO_(3))_(2)`D. `Sr(NO_(3))_(2)` |
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Answer» Correct Answer - D `Sr(NO_(3))_(2) overset(Delta) SrO + NO_(2) + O_(2)` |
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| 25889. |
Uterus is related to(A) Male Reproductive system(B) Female Reproductive system(C) Plant Reproductive system(D) All of these |
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Answer» Answer (B) Female Reproductive system |
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| 25890. |
Uracil is related to(A) RNA(B) DNA(C) Both (A) & (B)(D) None of these |
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Answer» Answer (A) RNA |
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| 25891. |
Organic evolution was preceded by chemical evolution, the champions of this theory are (A) A.l.Oparin and J.B.S.Haldane (B) Charles Darwin (C) Arrhenius (D) Baptiste Lamarck |
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Answer» (A) A.l.Oparin and J.B.S.Haldane |
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| 25892. |
Uracil is related to- (A) RNA (B) DNA (C) Both (A) & (B) (D) None of theses |
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Answer» Uracil is related to RNA. |
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| 25893. |
Synthesis of RNA on DNA template is known as- (A) Translation (B) Transcription (C) Transduction (D) Replication |
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Answer» (B) Transcription |
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| 25894. |
Climax community is present in which area? (A) In equilibrium (B) In transition (C) Baer land (D) None of these |
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Answer» (B) In transition |
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| 25895. |
Yeast reproduces by means of- (A) Budding (B) Fragmentation (C) Pollination (D) All of these |
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Answer» (A) Budding. |
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| 25896. |
Climax community is present in which area(A) In equilibrium(B) In transition(C) Bare land(C) None of these |
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Answer» Answer (A) In equilibrium |
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| 25897. |
Which of the following is a wrong pair ? (A) G.....C (B) T.....A (C) A ......U (D) T .....U |
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Answer» The Correct option is (D) T .....U |
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| 25898. |
Flowers of Vallisneria spp are(A) Anemophilous(B) Entomophilous(C) Hydrophilous(D) Zoophilous |
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Answer» Answer (C) Hydrophilous |
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| 25899. |
Yeast reproduces by means of(A) Budding(B) Fragmentation(C) Pollination(D) All of these |
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Answer» Answer (A) Budding |
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| 25900. |
Uterus is related to-(A) Male Reproductive system (B) Female Reproductive system (C) Plant Reproductive system (D) All of these |
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Answer» (B) Female Reproductive system |
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