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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.
| 97651. |
Consider the reaction: `2S_(2)O_(3)^(2-)(aq)+I_(2)(s) rarr S_(4)O_(6)^(2-)(aq) + 2I^(Θ)(aq)` `2S_(2)O_(3)^(2-)(aq) + 2Br_(2)(l) + 5H_(2)O(l) rarr 2SO_(4)^(2-)(aq) + 4Br^(Θ)(aq)+10H^(o+)(aq)` Why does the same reducatnt, thiosulphate, react differently with iodine and bromine? |
| Answer» The average oxidation number of `S` in `S_(2)O_(3^(2-)` is `+2` while in `S_(4)O_(6)^(2-)` it is `+2.5`. The oxidation number of `S` in `SO_(4)^(2-)` is `+6`. since `Br_(2)` is a stronger oxidising agent that `I_(2)`. It oxidies `S` to `S_(2)O_(3)^(2-)` to a higher oxidation state of `+6` and hence forms `SO_(4)^(2-)` ion. `I_(2)`, however, being a weaker oxidising agent oxidises `S` of `S_(2)O_(3)^(2-)` ion to a lower oxidation of `+2.5` in `S_(4)O_(6)^(2-)` ion. It is because of this reason thai thiosulphate reacts differently with `Br_(2)` and `I_(2)`. | |
| 97652. |
While sulphate dioxide and hydrogen perxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why? |
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Answer» a. In `SO_(2)`, oxidation number of `S` is `+4`. In principle `S` can have a minimum oxidation number of `-2` and maximum of `+6`. Therefore, `S` in `SO_(2)` can either decrease or increase its oxidation number and hence can act both as an oxidising as well as a reducing agent. b. In `H_(2)O`, the oxidation number of `O` is `-1`. In principle, `O` can can have a minimum oxidation number of `-2` and maximum of zero `(+2 " is possible only with " OF_(2))`. Therefore, `O` in `H_(2)O_(2)` can either decrease its oxidation number from `-1` to `-2` or can increase its oxidation number from `-1` to `0`. Therefore, `H_(2)O_(2)` acts as an oxidising as well as a reducing agent. c. In `O_(3)` the oxidation number of `O` is zero. It can only decrease its oxidation number from from `0` to `-1` or `-2`, but cannot increase to `+2`. Therefore, `O_(3)` acts only as an oxidant. d. In `HNO_(3)`, oxidation number of `N` is `+5` which is maximum. Therefore, it can only decrease its oxidation number and hence it acts as an oxidant only. |
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| 97653. |
How many types of polypeptides are found in intermediate filaments?(a) 2(b) 5(c) 10(d) 12 |
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Answer» The correct option is (b) 5 The best I can explain: Depending upon the cell type, intermediate filaments are composed of different polypeptide subunits. The polypeptide subunits found in intermediate filaments are divided into 5 major subclasses. |
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| 97654. |
In a mammalian cell, which is the most abundant type of RNA?(a) rRNA(b) mRNA(c) siRNA(d) tRNA |
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Answer» Correct choice is (a) rRNA The best I can explain: Ribosomal RNA (called rRNA) is the most abundant type of RNA in a cell. More than 80 percent of the RNA in a cell consists of ribosomal RNA. Other types of RNAs are transfer RNA and messenger RNA among others. |
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| 97655. |
Consider the following standard electrode potentials and calculate `logK_(eq)` at `25^(@)C` for the indicated disproportion reaction. [Take `(2.303RT)/F=0.06V`] `3MN^(2+)(aq)arrMn(s)+2Mn^(3+)(aq)` `Mn^(3+)(aq)+e^(-)rarrMn^(2+)(a),E^(@)=1.51V` `Mn^(2+)(aq)+2e^(-)rarrMn(s),E^(@)=-1.185V`A. `-21.15`B. `-48.24`C. `-89.83`D. none of these |
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Answer» Correct Answer - C `2Mn^(2+)rarr2Mn^(3+)+2e^(-),DeltaG_(1)^(0)` `Mn^(2+)+2e(-)rarrMn, DeltaG_(2)^(0)` `3Mn^(2+)(aq)rarrMn(s)+2Mn^(3+)(aq)` `=2xxFxxE_(3)^(0)=-2xxF[-1.51]-2xxFxx(-1.185)` `E_(3)^(0)=-2.695V` `E_(3)^(0)=+(0.06)/2logK_(eq)` |
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| 97656. |
In mammalian cells, the signal recognition particle (SRP) consists of ____ distinct polypeptides and one small RNA molecule.(a) 2(b) 4(c) 6(d) 8 |
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Answer» Correct answer is (c) 6 Easiest explanation: The hydrophobic signal present in the nascent polypeptide is recognized by the SRP which then binds to the SRP receptor and transports the peptide to the endoplasmic reticulum membrane. The signal recognition particle (SRP) consists of 6 distinct polypeptides and one small 7S RNA. |
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| 97657. |
A correct electrochemical series (increasing SRP) can be obtained from K. Li, Zn, Fe, H, Ag, Cu, Au by interchanging:A. K and LiB. Zn and FeC. Ag and CuD. Fe and H |
| Answer» Correct Answer - A::C | |
| 97658. |
A carbon resistor of `(47 +- 4.7) k Omega` is to be marked with rings of different colours for its identification. The colour code sequence will beA. Yellow- Green-Violet-GoldB. Yellow-Violet-Orange-SilverC. Violet-Yellow-Orange-SilverD. Green-Orange-Violet-Gold |
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Answer» Correct Answer - B Given, `R=(47 +- 4.7)kOmega = 47 xx 10^(3) +- 10%Omega` As per the color code for carbon resistors, the colour assigned to numbers. 4- yellow 7- Violet 3-Orange. For `+-10%` accuracy, the color is silver. Hence, the bands of colours on carbon resistor in sequence are yellow, violet, orange and silver. Note: To remember the colour code sequence for carbon resistor, the following sentenses should be kept in memory. B B Roy of Great Britian has a Very Good Wife. |
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| 97659. |
Which of the following galvanic cells no liquid junction potential?A. `Zn underset((C_(1)))(-)Hg|Zn-SO_(4)"solution"|Zn underset((C_(2)))(-)Hg`B. `H_(2) underset((C_(1)))(|)H_(2)SO_(4)"solution"||CuSO_(4)underset((C_(2)))("solution")|Cu`C. `underset((C_(1)))(Ag|AgCl|HCl"solution"|)underset((C_(2)))(|HCl "solution"|AgCl|Ag`D. none of these |
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Answer» Correct Answer - A Since cathode and anode are present in the same solution hence liquid -liquid junction potential is elimenated in option (A). |
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| 97660. |
A bag is gently dropped on a conveyor belt moving at a speed of 2 m/s. The coefficient of friction between the conveyor belt and bag is 0.4 Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion is : [Take g = 10 m/s–2](A) 2 m(B) 0.5 m(C) 3.2 m(D) 0.8 ms |
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Answer» Correct option is (B) 0.5 m In frame of belt a = µ g = 4 m/s2, v = 2m/s , u = 0 v2 = u2 + 2as ⇒ s = 0.5 m |
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| 97661. |
What is the two dimensional coordination number of a molecule in square close packed layer? |
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Answer» In square close-packed layer, a molecule is in contact with four of its neighbours. Therefore, the two-dimensional coordination number of a molecule in square close-packed layer is 4. |
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| 97662. |
Some of the very old glass objects appear slightly milky instead of being transparent. Why ? |
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Answer» Reallignment of molecules takes place due to movement because of constant heating and cooling. |
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| 97663. |
Calculate the mole fraction, molality and molarity of `HNO_(3)` solution contaning `12.2% HNO_(3)` .Given density of `HNO_(3) = 1.038 g cm^(-3)` |
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Answer» Mole fraction =0.03817 Molality = 2.205 mol `kg^(-1)` Molarity =2.01 M |
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| 97664. |
Calculate the number of atoms present in 2 grams of crystal which has face- centred cubic (FCC) crystal lattice having edge length of 100 pm and density 10 g` cm^(-3)` |
| Answer» Number of atoms `= 8 xx 10^(23)` | |
| 97665. |
Calculate enthalpy change for the reaction `C(dimond)+O_(2)(g) to CO_(2)(g)` Given: Energy required to break C-C bond in diamond is 350 kJ `mol^(-1)` `DeltaH_(f,O(g))^(@)=250kJ mol^(-1)` `DeltaH_("atomisation,CO_(2)(g)")^(@)=1500kJ mol^(-1)`A. `-300 kJ mol^(-1)`B. `-390 kJ mol^(-1)`C. `-400 kJ mol^(-1)`D. `-350 kJ mol^(-1)` |
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Answer» (A) `DeltaH^(@)=sum(BE)_(R )-sum(BE)_(P)` `=(2xx350+500)-(1500)` `=-300 kJ mol^(-1)` |
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| 97666. |
Why do the window glass of old old building (a) look milky and (b) become thick at the bottom? |
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Answer» a. Due to heating during the day and cooling at night, i.e., annealing over a number of years, glass acquires some crystalline character. b. Glass is not a true solid but a super cooling liquid of high viscosity (called pseudosolid). Therefore, it has the property to flow (i.e., fluidity). |
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| 97667. |
Naturally occurring gold crystallises in face - centred cubic structure and has density `19.3 g cm^(-3)` , Find atomic radius of gold `. (Au= 137 g "mol"^(-1))` |
| Answer» Correct Answer - 144 pm | |
| 97668. |
Give the name of one solid which shows both Schottky and Frenkel defects? |
| Answer» Correct Answer - `AgBr` | |
| 97669. |
Why stoichiometric defects are also called intrinsic defects? |
| Answer» Stoichiometric defects are so called because they do not alter the stochiometry of the crystal in both Schottky and Frenkel defects. They are called intrinsic defects because it is due to the deviation from regular arrangement of constituent particles with in the crystal and no external substance is added. | |
| 97670. |
Why amorphous solids are called as supercooled liquids? |
| Answer» Like liquids they are isotropic and process fluidity. | |
| 97671. |
Why the defectsof the crystalline solids are called thermodynamic defects? |
| Answer» The constituent particles in a crystal are in perfect order at `0 K`. With the increase of temperature, the chance that a lattice site may be unoccupied by an ion increases. Since the number of defects increases with temperature, the defects are called thermodynamic defect. | |
| 97672. |
If `NaCI` is droped with `10^(-3)` mole of `SrCI_(2)` then number of cationic vacancies is |
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Answer» The number of cation valencies in the lattice of `NaCl` is equal to the number of divalent `Sr^(2+)` ions added. The concentration of `Sr^(2+)` ion `= 10^(-3) mol` percent `(10^(-5) "mol")/(100) = 10^(-5) mol` ` :.` mol of `Sr^(2+)` ions contains `6.023 xx 10^(23) Sr^(2+)` ions `:. 10^(-5)` mol of `Sr^(2+)` ions contains `= (6.023 xx 10^(23)) xx 10^(-5) = 6.023 xx 10^(18) Sr^(2+)` ions Therefore, the number of cation vacancies in `NaCl` crystal is `6.023 xx 10^(18)`, when it is doped with `10^(-3)` mole% of `SrCl_(2)` |
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| 97673. |
Example the following with suitable examples: a. Ferromagnetism b. Paramagnetism c. Ferrimagnetism d. Antiferromagnetism e. 12 - 46 and 13 - 15 group compounds |
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Answer» 12-16 and 13-15 groups compounds i. A solid binary compound by combining an element of group 12 with an element of group 16 is called 12-16 compound. `ZnZ, CdS, HgTe` are 12-16 compounds. ii. A solid binary of an element of group 13 and an element of group 15 is called 13-15 compound. `AIP, GaAs`, and `InSb`, are example of 13-15 compounds. |
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| 97674. |
What type of stoichiometric defect is shown by: (i) ZnS (ii) AgBr |
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Answer» (i) ZnS shows Frenkel defect. (ii) AgBr shows Frenkel defect as well as Schottky defect. |
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| 97675. |
What type of stoichiometric defect is shown by:i. ZnSii. KCliii. AgBr |
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Answer» i. ZnS shows Frenkel defect due to a large difference in size of Zn2+ ion and S2- ion. |
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| 97676. |
What is the minium temperature of MgO by carbon ?A. below 1600 KB. above 1600 KC. 1600 KD. 1873 K |
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Answer» Correct Answer - B Since above 1600 K , `DeltaG^(@)` for the formation of `CO_(2)` is more negative than for the formation of MgO, the minimum temperature required for the reduction of MgO by carbon is 1600 K. |
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| 97677. |
The flowchart of the process of concentrate of aluminium ore is given. Complete the flowchart. |
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Answer» a. NaAlO2 (Sodium aluminate) b. Al(OH)3 /Aluminum hydroxide) c. The precipitate is separated, washed well and strongly heated; d. Al2O3 |
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| 97678. |
Two compounds of iron are jpven below. FeSO4 Fe2(SO4)3 (The oxidation state of sulfate radical is-2) a. Which of these compounds show +2 oxidation state for Fe? b. Which compounds has Fe3+ ion? c. Write the subshell electronic configuration of Fe3+ ion. d. Why do transition elements show variable oxidation states? |
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Answer» 1. a. FeSO4 b. Fe2(SO4)3 c. Fe3+ – 1s2 2s2 2p6 3s2 3p6 3d5 d. The energy difference between the outer most ‘s’ subshell and penultinate ‘d’ subshell is very small. Hence under suitable conditions, the electrons in ‘d’ subshell also take part in chemical reaction. |
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| 97679. |
A body of mass `4 kg` moving with velocity `12 m//s` collides with another body of mass `6 kg` at rest. If two bodies stick together after collision , then the loss of kinetic energy of system isA. zeroB. 288JC. 172.8 JD. 144 J |
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Answer» Correct Answer - C |
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| 97680. |
The Coefficient of cubical expansion of mercury is `0.0018//^(@)C` and that of brass `0.000006//^(@)C`, find the true barmetric height at `0^(@)C`.The scale is supposed to be coreect at `15^(@)C`A. 74.122 cmB. 79.125 cmC. 42.161cmD. 142.39 cm |
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Answer» Correct Answer - A |
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| 97681. |
A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration `a_(c)` is varying with time t as `a_(c) = k^(2)rt^(2)`, where k is a constant. The power delivered to the particle by the forces acting on it is :A. `2pimk^(2)r^(2)t`B. m`mk^(2)r^(2)t`C. `1//3mk^(4)r^(2)t_(5)`D. zero |
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Answer» Correct Answer - B |
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| 97682. |
A metal rod lenth `L_(0)` whose coefficient of linear expansion `alpha = 10^(-3) ^(0)C^(-1)` is heated such that its temperature changes by `1000K` assuming `alpha` is constant during the temperature change `(e =2.7)`.Which of the following statements are trueA. Final length of the rod is greater than 2L.B. Final length of the rod is greater than 2.5 LC. Final length of the rod is greater then 3LD. increase in the length due to heating is L |
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Answer» Correct Answer - A::B `alpha=(1)/(L)(dL)(dT)` `int10^(-3)dT=int(dL)/(L)` `L_(f)=2.7L` |
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| 97683. |
A liquid rises to a height of 1.8 cm in a glass capillary A another glass capillary B having diameter 90% of capillary A is immersed in the same liquid the rise of liquid in capillary B isA. 1.4 cmB. 1.8 cmC. 2.0 cmD. 2.2 cm |
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Answer» Correct Answer - a Height of liquid in a capillary tube , `h =(2S cos theta)/(rpg)` where S= surface tension p = density of the liquid r= radius of ht capillary tube or `(1.8 cm)=h_(B)xx9/10` or `h_(B)=10/9xx1.8 cm =2cm` |
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| 97684. |
If 10g of A(molar mass = 20g) reacts with 30g of B(molar mass = 30 g) according to following steps. Step-I `A+3BrarrC+2D` Step-II `D+2Erarr5F` Calculate the produced mole of F, if % yield of Ist and IInd steps are 100 & 60 respectively |
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Answer» Correct Answer - 2 `{:(,A,+,3B,rarr,C+2D,),("Mass",10g,,30g,,,),("Mole",(1)/(2),,1,,,),("Mole S.C.",(1)/(2),,(1)/(3),,,):}` Produced moles of `D = (2)/(3) xx 1` `therefore` Produced moles of `F = (5)/(1) xx(2)/(3)xx(60)/(100)=2` |
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| 97685. |
A small ball of mass m hits the cylinder which is hinged at top and free to rotate in vertical plane as shown in figure. Mass of cylinder is M. Linear momentum of system (ball+cylinder)A. Remains conserved if y=RB. Remains conserved if `y=(3)/(2)R`C. Remains conserved if `y=(7)/(5)R`D. Cannot be conserved for any value of y. |
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Answer» Correct Answer - B muy`=mVy+(3)/(2)MR^(2)omega` where u is speed of ball before collision, V speed of ball after collision and `omega` is angular speed of cylinder after collision. `m u=mV+Mromega(COM)` `thereforey=(3R)/(2)` |
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| 97686. |
A slender rod is hinged at its top end and hanging freely in a vertical position. A tiny ball, moving horizontally, hits the lower end of the rod elastically and comes to rest. The ratio of the mass of rod to the mass of the ball must beA. `3:2`B. `3:1`C. `5:3`D. insufficient data |
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Answer» Correct Answer - 2 `mv_(0)l=(ml^(2))/3 omega` `v_(0)=omegalrArr M=3m` |
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| 97687. |
The apparent coefficient of expansion of liquid, when heated in a copper vessel is `C` and when heated in a silver vessel is `S`. If `A` is the linear coefficient of expansion of Copper, linear expansion coefficient of silver isA. `(C+S-3A)/(3)`B. `(C+3A-S)/(3)`C. `(S+3A-C)/(3)`D. `(C+S+3A)/(3)` |
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Answer» Correct Answer - B `{:(y_(a)=y_(r)-y_(c)),(C=y_(R)-3A),(S=y_(R)-3alpha_(s)),(C-S=3alpha_(s)-3A):}` `alpha_(s)=(C-S+3A)/(3)` |
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| 97688. |
Two soap bubbles, each of radius r, coaleses in vacuum under isotermal conditions to from a bigger bubble of radius R. Then R is equal toA. `2^(-1//2)r`B. `2^(-1//3)r`C. `2^(1//2)r`D. 2r |
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Answer» Correct Answer - C `4pir_(1)^(2)+4pir_(2)^(2)rArr4piR^(2)` `R=sqrt(2r)` |
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| 97689. |
If the volume of a block os aluminium is decreased be `1%` the pressure (stress) on is surface is increased by (Bulk moduals) of `Al=7.5xx10^(10)Nm^(-2))`A. `7.5xx10^(10)Nm^(-2)`B. `7.5xx10^(8)Nm^(-2)`C. `7.5xx10^(6)Nm^(-2)`D. `7.5xx10^(4)Nm^(-2)` |
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Answer» Correct Answer - B Given, `(DeltaV)/(V)=1%=(1)/(100)` Bulk modulus, `B=(P)/(DeltaV//V)=(PV)/(DeltaV)` or `P=(BDeltaV)/(V)=7.5xx10^(10)xx(1)/(100)` `=7.5xx10^(8)Nm^(-2)` |
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| 97690. |
Statement 1: True solution and colloidal solution can be separated using parchment paper.Statement 2: Particles of true solution does not pass through parchment paper whereas colloidal solution passes.(a) Both statement 1 and 2 are correct(b) Statement 1 is correct but statement 2 is incorrect(c) Statement 1 is incorrect but statement 2 is correct(d) Both statement 1 and 2 are incorrect. |
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Answer» (d) Both statement 1 and 2 are incorrect. True solution cannot be separated using parchment paper Particles of colloidal solution cannot pass through parchment paper Therefore, both the statements are false |
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| 97691. |
In an adsorption experiment a graph between log `x/m` versus log `P` was fond to be linear with a slope of `45^@)`. The intercept on the log `x/m` axis was found to `0.70`. Calculate the amount of gas adsorbed per gram of charcoal under a pressure of `1` atm `[log5=0.70]`A. `5`B. `2`C. `3`D. `0` |
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Answer» Correct Answer - A `"log" x/m=logk+1/nlogP` `1/n=tan45^(@)` `:.n=1` `logk=0.70` `:.k=5` Now `x/m=KP^(1//m)` `5(1)^(1/1)=5` |
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| 97692. |
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).Assertion (A): Dissolved substances can be removed from a colloidal solution by diffusion through a parchment paper.Reason (R): Particles in a true solution cannot pass through parchment paper but the collodial particles can pass through the parchment paper.In the light of the above statements, choose the correct answer from the options given below:(A) Both (A) and (R) are correct and (R) is the correct explanation of (A)(B) Both (A) and (R) are correct but (R) is not the correct explanation of (A)(C) (A) is correct but (R) is not correct(D) (A) is not correct but (R) is correct |
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Answer» Correct option is (C) (A) is correct but (R) is not correct Assertion (A): Correct. Reason(R): Incorrect. Particles of true solution pass through parchment paper thus answer is (C). |
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| 97693. |
Assertion: Dissolved substance of colloidal solution can be separated by using parchment paper.Reason: Particles of true solution can not pass through parchment paper while particles of colloidal solution pas through parchment paper.(1) Assertion is true, Reason is true and Reason is correct explanation of Assertion(2) Assertion is true, Reason is true and Reason is not correct explanation of Assertion.(3) Assertion is true and Reason is false(4) Assertion is false and Reason is true |
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Answer» Correct option is (3) Assertion is true and Reason is false Colloidal solution particles can be separated using parchment paper. True solution can pass through parchment paper while colloidal particle can not pass. |
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| 97694. |
Distinguish between multimolecular, macromolecular and associated colloids with the help of one example of each. |
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Answer» [Hint : (i) Multimolecular colloids formed by aggregation of small atoms or molecules. Example, gold sol, sulphur sol. |
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| 97695. |
State "Hardy Schulze Rule' with one example. |
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Answer» [Hint : It states that greater the valency of the occulating ion of the electrolyte, the faster is the coagulation.] |
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| 97696. |
If the doctor gives you 12 pills tells you to take 4 pills every half hour, like this how long will it take you to finish all those pills?1. 2 hours2. 2 hours 15 Mins3. 1 Hour 30 Mins4. 2 hours 30 Mins |
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Answer» Correct Answer - Option 3 : 1 Hour 30 Mins Given: Total number of pills = 12 Frequency to take pills = 4 pills per hour Calculation: Time taken to finish 4 pills = 30 minutes Then, time taken to finish 12 pills = (30/4) × 12 minutes ⇒ 90 minutes = 1 hour 30 minutes ∴ Time taken to finish 12 pills is 1 hour 30 minutes |
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| 97697. |
Name the impurities present in bauxite ore. |
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Answer» SiO2, Fe2O3 and TiO2. |
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| 97698. |
Assertion(A): The micelle formed by sodiumm stearate in water has -COO groups at the surface. Reason(R): Surface tension of water is reduced by addition of stearate.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If assertion is false but reason is true. |
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Answer» Correct Answer - B Micelle is formed if molecules with polar and non-polar ends assemble in bluk to give non-polar interior and polar exterior. |
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| 97699. |
On adding Kl to a metal salt solution, no precipitate was observed but the salt solution gives yellow precipitate with `K_(2)CrO_(4)` in the presence `CH_(3)COOH`. Then the salt isA. `Sr(NO_(3))_(2)`B. `Pb(CH_(3)COO)_(2)`C. `AgNO_(3)`D. `BaCl_(2)` |
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Answer» Correct Answer - D `BaCl_(2)+KltoNo" ppt"` `BaCl_(2)(aq)+CrO_(4)^(2+)tounderset("yello ppt")(BaCrO_(4)darr` |
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| 97700. |
A freshly prepared `Fe(OH)_(3)` precipitate is peptized by adding `FeCl_(3)` solution. The charge on the colloidal particle is due to preferential adsorption ofA. `Cl^(-)`B. `Fe^(3+)`C. `OH^(-)`D. none of these |
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Answer» Correct Answer - B Preferential adsorption of `Fe^(3+)` takes plaec so positive charge develop |
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