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InterviewSolution
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.
| 1351. |
The number of octahedral sites per sphere in fcc structure isA. `8`B. `4`C. `2`D. `1` |
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Answer» Correct Answer - D Number of octahedral sites `=` Number of ions in the packing. `:.` Number of octahedral sites per sphere `= 1` |
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| 1352. |
The number of octahedral sites per sphere in fcc structure isA. 1B. 2C. 4D. 8 |
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Answer» Correct Answer - 1 In any close packing , no of spheres is equal to no . Of octahedral holes. |
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| 1353. |
Assertion. In a unit cell of NaCl, all ` Cl^(-)` ions as will they touch each other. Reason. Radius ratio ` r_(+)//r_(-)` in NaCl is 0.414. |
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Answer» Correct Answer - a Correct A. In a unit cell of NaCl, all ` Cl^(-)` ion touch ` Na^(+)` ion but do not touch each other. Correct R. Actual radius ratio ` r_(Na^(+))//r_(Cl^(-))` is 0.525 and not 0.414. |
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| 1354. |
First order X-ray reflection `(lambda=154pm)` is maximum from a set of (2,0,0) planes of a bcc lattice was observed at `16^(0)6^(1)`. Then the edge length of unit cell is `(Sin 16^(0)6^(1)=0.2773)`A. 360pmB. 420pmC. 560pmD. 180ppm |
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Answer» For first order reflection, n=1 `lambda=2d sin theta` `d=(lambda)/(2 sin theta)=(154)/(2xx0.2773)=280` pm `d_(((2,0,0)))=(alpha)/sqrt(2^(2)+0^(2)+0^(2))rArr a=2xx280=560` pm |
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| 1355. |
The concentration of cation vacancies per mole when NaCl is doped with `10^(-3)` mole % of `SrCl_(2)`A. `6.023xx10^(20)`B. `6.023xx10^(23)`C. `6.023xx10^(21)`D. `6.023xx10^(18)` |
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Answer» no. of moles of `Sr^(+2)` added `=10^(-3)xx(1)/(100)=10^(-5)` moles no. of `Sr^(+2)` ions added `=6.023xx10^(23)xx10^(-5)=6.023xx10^(18)` |
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| 1356. |
Assertion. Size of the cation is larger in tetrahedral hole than in an octahedral hole. Reason. The cations occupy more space than anions in crystal packing. |
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Answer» Correct Answer - d Correct A. Octahedral hole is large in size than tetrahedral hole. Correct R. cations are generally smaller. They occupy the voids and hece occupy less space. |
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| 1357. |
In a close-packed structure of oxides, one-eighth of the tetrahedral holes is occupied by bivalent cations `(A)` and half of the octahedral hole are occupied by trivalent cations `(B)`, calculate the molecular formula of the oxide. |
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Answer» Correct Answer - `AB_(2)O_(4)` Let number of `O^(2-) = 1//"atom"` Number of `TV = 2//"atom"` Number of `TV = 1//"atom"` Number of `A^(2+) = (1)/(8) xx 2 =(1)/(4)` Number of `B^(3+) = (1)/(2) xx 1 =(1)/(2)` Formula: `A_(1//4)^(2+)B_(1//4)^(3+)O^(2-)` or `AB_(2)O_(4)`. |
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| 1358. |
A solid compound `XY` has `NaCl` structure. If the radius of the cation is `100` pm, the radius of the anion `(Y^(-))` will beA. 241.5pmB. 165.7pmC. 275.1pmD. 322.5pm |
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Answer» Correct Answer - A Radius ratio of NaCl like crystal `=(r^(+))/(r^(-))=0.414` |
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| 1359. |
Define radius ratio. What is the value of radius ratio for octahedral geometry ? |
| Answer» For octahedral geometry, radius ratio is in the range `0.414-0.732`. | |
| 1360. |
What type of semi-conductor is produced when silicon is doped with boron ? |
| Answer» Silicon has 4 valence electrons while boron has 3 valence electrons. Dopping silicon with boron leads to positive hole without any electron. This results in p-type semiconductors. | |
| 1361. |
Dopping of AgCl crystals with `CdCl_(2)` results in:A. Schottky defectB. Frenkel defectC. Substitutional cation vacancyD. Formation of F-centres |
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Answer» Correct Answer - C |
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| 1362. |
What change occurs when AgCl is doped with `CdCl_(2)` ? |
| Answer» Some `Ag^(+)` ions are replaced by `Cd^(2+)` ions due to smaller size. Actually to accommodate one `Cd^(2+)` ion, two `Ag^(+)` ions are to leave the lattice site in order to maintain electrical neutrality. One vacant position will be filled by `Cd^(2+)` ion while the other will remain vacant as a hole. This will lead to increased electrical conductivity of the solid. | |
| 1363. |
Assertion (A) : Group-`13`-doped crystals of `Si` are called `p`-type semiconductors. Reason (R ) : Positive holes are reasponsible for the semiconducting properties.A. If both `(A)` and `(R)` are correct, and `(R)` is the correct explanation of `(A)`B. If both `(A)` and `(R)` are correct, but `(R)` is not the correct explanation of `(A)`C. If `(A)` is correct, but `(R)` is incorrect.D. If both `(A)` is incorrect, but `(R)` is correct. |
| Answer» Correct Answer - A | |
| 1364. |
Assertion (A) : Group-`13`-doped crystals of `Si` are called `p`-type semiconductors. Reason (R ) : Positive holes are reasponsible for the semiconducting properties.A. If both assertion and reason are correct and reason is correct explanation for assertionB. If both assertion and reason are correct but reason is not correct explanation for assertionC. If assertion is correct but reason is incorrectD. If both assertion and reason are incorrect |
| Answer» Correct Answer - A | |
| 1365. |
(a) What type of semiconductor is obtained when silicon is doped with boron ? (ii) What type of magnetism is shown in the following alignment of magnetic moments ?`uarr uarr uarr uarr ` What type of point defect is produced when AgCl is doped with `CdCl_(2)` ? |
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Answer» Correct Answer - (a) SiC = covalent/network solid, Ar = Non-polar mollecular solid (b) N/A |
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| 1366. |
The oxide that is insulator isA. VOB. coOC. `ReO_(3)`D. `Ti_(2)O_(3)` |
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Answer» Correct Answer - B Oxide that is insulator is CoO. |
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| 1367. |
Assertion (A) : Group-`13`-doped crystals of `Si` are called `p`-type semiconductors. Reason (R ) : Positive holes are reasponsible for the semiconducting properties.A. If both assertion and reason are correct and reason is correct explanation for assertionB. If both assertion and reason are correct but reason is not correct explanation for assertionC. If assertion is correct but reason is incorrectD. If both assertion and reason are incorrect |
| Answer» Correct Answer - A | |
| 1368. |
Classify each of the following as being n-type semi-conductor. Give reason. (a) Si doped with In (b) Si doped with R. |
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Answer» (a) It is p-typed semi-conductor. (b) It is n-type semi-conductor. |
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| 1369. |
The n-type semiconductor is obtained when Si is doped withA. AlB. GeC. BD. As |
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Answer» Correct Answer - D n-type semicondutor (negative) is formed when Si is doped withAs, due to extra electron in valence shell of As which favours conduction |
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| 1370. |
If Si is doped with B,A. n-type semiconductorB. p-type semiconductorC. a combination of the above two typesD. None of the above |
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Answer» Correct Answer - B A/c to definition |
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| 1371. |
Which onr of the following statements is wrong ?A. The conductivity of metals decreases with increase in temperatureB. The conductivity of semiconductors increases with increase in temperatureC. There is no superconductor at room temper -tureD. Ionic solids conduct electricity due to presence of ions . |
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Answer» Correct Answer - D Ionic solids conduct electricity not due to presence of ions but due to presence of defects . |
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| 1372. |
In a solid ,oxide `(O^(2-))` ions are arranged in ccp, cations `(A^(3+))` occupy one -sixth of tetrahedral void and cations `(B^(3+))` occupy one -third of the octahedral voids . What is the formula of the compound? |
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Answer» Correct Answer - `ABO_(3)` In ccp with each oxide, there would be 2 tetrahedral voids and one octahedral void. According to available data, `1//3^(rd)` octahedral voids are occupied by B and 1/6th tetrahedral voids by A. Therefore, the formula of the compound is `A_(2//6)B_(1//3)O or A_(2)B_(2)O_(6) or ABO_(3)`. |
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| 1373. |
STATEMENT -1 : In NaCl structure , ` Na^(+)` ion occupy octahedral holes and `Cl^(-)` ions occupy ccp. STATEMENT -2 : The distance of the nearest neighours in NaCl structure is `a//2` where a is the edge length of the cube .A. If both the statement are TRUE and STATEMENT -2 is the correct explanation of STATEMENT -1B. If both the statement are TRUE and STATEMENT -2 is NOT the correct explanation of STATEMENT -1C. If STATEMENT -1 is the correcct and TRUE and STATEMENT -2 is FALSED. If STATEMENT -1 is the correcct and FALSE and STATEMENT -2 is TRUE |
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Answer» Correct Answer - B |
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| 1374. |
Tungsten has an atomic radius of 0.136nm. The density of tungsten is `19.4g//cm^(3)`. What is the crystal structure of tungsten ? `("Atomic mass" W=184)`A. simple cubicB. Body centred cubicC. Face centred cubicD. none of these |
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Answer» Correct Answer - B |
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| 1375. |
Statement-1 C.N of `Cs^(+)` ion in CsCl structure is 8 Statement -2 CsCl crystallizes in BBC structureA. If both the statement are TRUE and STATEMENT -2 is the correct explanation of STATEMENT -1B. If both the statement are TRUE and STATEMENT -2 is NOT the correct explanation of STATEMENT -1C. If STATEMENT -1 is the correcct and TRUE and STATEMENT -2 is FALSED. If STATEMENT -1 is the correcct and FALSE and STATEMENT -2 is TRUE |
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Answer» Correct Answer - D |
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| 1376. |
STATEMENT -1 : For fluorite structure , the `F^(-)` ions occupy tetrahedral void and `Ca^(2+)` ions in ccp STATEMENT-2 : The radius ratio of fluorte structure is 0.414A. If both the statement are TRUE and STATEMENT -2 is the correct explanation of STATEMENT -1B. If both the statement are TRUE and STATEMENT -2 is NOT the correct explanation of STATEMENT -1C. If STATEMENT -1 is the correcct and TRUE and STATEMENT -2 is FALSED. If STATEMENT -1 is the correcct and FALSE and STATEMENT -2 is TRUE |
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Answer» Correct Answer - C |
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| 1377. |
The density of argon (face centered cubic cell) is `1.83g//cm^(3) at 20^(@)C`. What is the length of an edge a unit cell? `("Atomic mass": Ar=40)`A. 0.599nmB. 0.569nmC. 0.525nmD. 0.551nm |
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Answer» Correct Answer - C |
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| 1378. |
The face centered cubic cell of platinum ha an edge length of 0.392nm. Calculate the density of platinum `(g//cm^(3))` : `("Atomic mass": Pt=195)`A. 20.9B. 20.4C. 19.6D. 21.5 |
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Answer» Correct Answer - D |
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| 1379. |
Face centered cubic lattice of NaCl may also smaller, tetragonal unit cell as shown below fig. (a,b) The number of formula units of NaCl present in the smaller unit cell drawn in (a) areA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B B, Unit cell (a): the slating unit cell length is `(l)/sqrt(2)`, where l=length of fcc unit cell. Unit cell (b): the angle between the two edges on the front face (heavier lines) is `90^(@)` . This unit cell extends only `(l)/(2)` towards back, there fore the volume of unit cell `(l)/sqrt(2)xx(l)/sqrt(2)xx(l)/sqrt(2)=(l^(3))/(4)=(v)/(4)` |
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| 1380. |
In a crystalline solid, atoms of X form fcc packing and atoms of Y occupy all octahedral voids. If all the atoms along one body diagonal are removed, what is the simplest formula of the crystalline lattice solid? |
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Answer» Correct Answer - `X_(5)Y_(4)` Number of atoms of X in fcc packing `=8xx(1)/(8)+6xx(1)/(2)=4` Number of atoms of Y at octahedral voids = 4 Along one body diagonal, there are two X atoms and one Y atom. Number of effective atoms of X after removal = `4-2xx(1)/(8)=(15)/(4)` Number of effetive atoms of Y after removal = 4-1=3 `X:Y=(15)/(4):3=5:4` Simplest formula of the solid `=X_(5)Y_(4)`. |
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| 1381. |
A crystal is made up of particals `X`, `Y`, and `Z`. `X` froms `fcc` packing. `Y` occupies all octahedral voids of `X` and `Z` occupies all tetrahedral voids of `X`. If all the particles along one body diagonal are removed. Then the fromula of the crystal would beA. `XYZ_(2)`B. `X_(2)YZ_(2)`C. `X_(8)Y_(4)Z_(5)`D. `X_(5)Y_(4)Z_(8)` |
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Answer» Correct Answer - D For fcc, number of `X` atoms `= 4//"unit cell"` Number of `TVs = 8Z` Number of `OVs = 4Y` Number of atoms removed along one body diagonal `= 2X ("corner")` and `2Z("TVs")` are `1Y` (OV at body centre) `:.` Number of `X` atoms left `= 4 - (2 xx 1/8) = 15/4` Number of `Y` atoms left `= 4 - (1 xx 1) = 3` Number of `Z` atoms left `= 8 - (2 xx 1) = 6` The simplest formula `= X_(15/4)Y_(3)Z_(6) rArr X_(15)Y_(12)Z_(24)` `rArr X_(5)Y_(4)Z_(8)` |
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| 1382. |
Calculate the packing effeciency of a fcc crystal in which all the tetrahedral and octahedral voids are occupied by the largest spheres without disturibing the lattice. |
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Answer» Fcc has 4 atoms per unit cell No. of tetrahedral voids = 8 No. of octadedral voilds = 4 If R is the radius of the atoms in the packing Radius of tetahedral void = 0.225 R Radius of octahedral void = 0.414 R for fcc, ` a = 2 sqrt2 r = 2sqrtR` packing efficiency ` = (" Volume occupied by all spheres")/("Volume of the unit cell") ` ` = (4 xx 4/3 pi R^(3) + 8 xx 4/3 pi ( 0.225 R)^(3) + 4 xx 4/3 pi ( 0.414 R)^(3))/((2sqrt2R)^(3))` = 0.81 |
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| 1383. |
Which of the folowings statements are not true ?A. Vacancy defect results in a decrease in the density of the substance.B. Intersitital defect results in an increase in the density of the substance .C. Impurity defect has no effect on the density of the substance.D. Frenkel defect results in an increase in the density of the substance. |
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Answer» Correct Answer - A::B::D © is not true because impurity defect changes the mass but not the volume. (d) is not true because Frenkel defect neither changes mass nor volume. |
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| 1384. |
Which of the following statements are true about metals ?A. Valence band overlaps with conduction band.B. The gap between valence band and conduction band is negligible.C. The gap between valence band and conducition band cannot be determined.D. Valance band amy remain partially filled. |
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Answer» Correct Answer - A::B Only © is not true. |
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| 1385. |
a. `MgO` has the structure of `NaCl` and `TICI` has the structure of `CsCl` . What are the coordiantion number of the ions in `MgO` and `TICI`? If the closed-packed cations in an `XY`-type solid have a raidus of `73.2` pm, what would be the maximum and minimum sizes of the anions filling voids? c. `Fe_(2)O_(3)` (haematite) forms ccp arrangement of `O^(2-)` ions with `Fe_(3+)` ions occupying initerstitial positions. Predict whether `fe^(3+)` ions are in the `OV` or `TV`. Given `r_(Fe^(3+)) = 0.7 Å` and `r_(O^(2-)) = 1.4 Å` d. A solid `XY` has `CsCl`-type structure. The edge length of the unit cell is `400 pm` Calculate the distance of closest approach between `X^(o+)` and `Y^(ɵ)` ions. |
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Answer» Correct Answer - C.N. of `Mg^(2+)&O^(2-)` ions =6, C.N. of `Cs^(+)&Cl^(-)` ions =8 C.N. of `Na^(+)` in NaCl = 6 , C.N of `Cl^(-)` in NaCl = 6. Hence in MgO C.N. of `Mg^(2+)` is also = 6 , that of `O^(2-)=6` We know that in CsCl C.N. pf `Cs^(+)=8` , C.N. of `Cl^(-)=8` Hence, `Ti^(+)` and `Cl^(-)`,in TiCl have also C.N. 8 each. |
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| 1386. |
What is the formula of a coumound in which the element Y forms ccp lattice and atoms X occupy 1/3rd of tetrahedral voids ? |
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Answer» Correct Answer - `X_(2)Y_(3)` Let the number of atoms of Y in ccp lattice = n Number of tetrahedral voids = 2n Since only 1/3 of tetrahedral voids are occupied by the atoms of the elements X Number of atoms of X in ccp lattice = `2nxx1//3=2n//3.` Ratio of X:Y is 2n/3 : n = 2 : 3 Formula of the compound = `X_(2)Y_(3)`. |
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| 1387. |
A compound formed by two elements `M` and `N`. Element `N` forms ccp and atoms of `M` occupy `1//3rd` of tetrahedral voids. What is the formula of th compound? |
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Answer» Let us suppose that, the no. of atoms of N present in ccp = x Since 1/3rd of the tetrahedral voids occupied by the atoms of M, therefore, the no. of tetrahedral void occupied = 2x/3 The ratio of atoms of N and M in the compound = `x:2x//3or 3:2` `therefore` The formula of the compound = `N_(3)M_(2)orM_(2)N_(3)` |
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| 1388. |
A compound formed by two elements `M` and `N`. Element `N` forms ccp and atoms of `M` occupy `1//3rd` of tetrahedral voids. What is the formula of th compound?A. `MN_(3)`B. `M_(3)N`C. `M_(3)N_(2)`D. `M_(2)N_(3)` |
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Answer» Correct Answer - D To solve such problems follow the following steps Setp I: Find the number of tetrahedral voids as number of tetrahedral voids `=2xx` number of atoms present in the lattice. Setp II: Calculate the number of atoms(or ratio ) of elements M and N as a chemical formula represents the number of atoms of different elements present in a compound. Derive the formula . Suppose atoms of element N present in ccp = x Then number of tetrahedral voids `=2xx` Since `(1)/(3)` rd of the tetrahedral voids are occupied by atoms of element M. Therefore, number of atoms of element `M=(1)/(3)xx2x xx(2x)/(3)` Ratio of `M:N=(2x)/(3):x=2:3` |
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| 1389. |
The energy gap ` (E_(g))` between valence band and conduction band for diamond , silicon and germanium are in the orderA. `E_(g) "( diamond)" lt E_(g) " (Silicon) " lt E_(g) " ( germanium) " `B. ` E_(g) " (diamond)" lt E_(g) "(silicon )" lt E_(g) " (germanium)" `C. `E_(g) " ( diamond)" = E_(g) " ( silicon)" = E_(g) " ( germanium) " `D. `E_(g) " ( diamond) " lt E_(g) " ( germanium ) " lt E_(g) " ( silicon) " ` |
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Answer» Correct Answer - a is The correct option because diamond is an insulator and non-metallic character decreases down the group. |
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| 1390. |
Copper crystallizes into a FCC lattice with edge length `3.61 xx 10^(-8)` cm . Show that the calculated density is in agreement with its measured value of `8.92 g. cm^(-3)` . |
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Answer» Density `d = (ZM)/(a^(3) xx N_(A))` Given edge length = `3.61 xx 10^(-8)` cm For FCC lattice of copper Z = 4 Atomic mass of copper M = 63.5 gms/mole `d = (4 xx 63.5)/((3.61 xx 10^(-8))^(3) xx 6.023 xx 10^(23))` `= 8.97 g//cm^(3)` The calculated value is approximately in agreement with the measured value `8.92 g// cm^(3)`. |
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| 1391. |
Ferric oxide crystallizes in a hexagonal close - packed array of oxide ions with two of every three octahedral holes occupied by ferric ions . Derive the formula of ferric oxide . |
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Answer» Given ferric oxide crystallises in a hexagonal close packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions . In hexagonal close - packed arrangement there is one octahedral hole for each atom . If the number of oxide ions `(O^-2)` per unit cell ,then the number of `Fe^(+3)` ions = `2 //3 xx` octahedral holes . `= 2//3 xx 1 = 2//3` The formula of the compound = `F_(2//3) O_(1)` (or) `Fe_(2) O_(3)` . |
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| 1392. |
Ferric oxide crystalliizes in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide. |
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Answer» There is one octahedral hole for each atom in hexagonal close packed arrangement. If the number of oxide ions `(O^(2-))` per unit cell is 1, then the number of `Fe^(3+)` ions `=2//3xx"octahedral holes"=2//3xx1=2//3`. Thus, the formula of the compound `=Fe_(2//3)O_(1)or Fe_(2)O_(3)`. |
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| 1393. |
(a) What is the C.N. of Cr atom in bcc structure ? (b) Cobalt metal crystallises in a hexagonal closest packed structure. What is the C.N. of cobalt atom? (c) Describe the crystal structure of Pt, which crystallises with four equivalent atoms in a cubic unit cell |
| Answer» (a) 8 (b) 12 (c) fcc or cubic close packed. | |
| 1394. |
In hexagonal closest-packed arrangement each spher has a coordinationnumber ofA. `12`B. `8`C. `4`D. `6` |
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Answer» Correct Answer - 1 In the hexagonal closest packed structure the first and thrid layers are oriented in the same direciton so that each atom in the third layer lies directly above an atom in the first layer. Each sphere has a coordination numberof `12`-six neighbors in the same layer .Three above and three below. |
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| 1395. |
In which of the following packing 74% space is occupied by the atoms with ABCABC packing of atoms ?A. Hexagonal closed packingB. Simple cubicC. Body centred cubicD. Cubic closed packing |
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Answer» Correct Answer - D In cubic closed packing , 74% space is occupied by the atoms with ABCABC packing of atoms , |
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| 1396. |
AB AB …….. Type of packing is called …… whereas ABCABC…….typeof packing is called …… |
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Answer» Correct Answer - 12,8 |
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| 1397. |
Hexagonal closet packed arrangement of equal -sized spheres is described byA. `ABCACB…...`B. `"ABBAAB"…..`C. `ABABAB…..`D. `ABCABABC….` |
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Answer» Correct Answer - 3 Each closest packed layer of spheres is formed by placeing spheres in the crevices at the adjoining row when a spher is placed in a hollow on top of a layer three other hollows are partially covered ,there are now two types of hollows on top of the second layer ,labeled `x` and `y` . The `x` sites are not . Sphere of a third layer may occupy either all of the sites or all of the `y` sites. When the spheres in the third layer are placed in the `x`-sites so that the third layer repeats the first layer we label this stacking `ABA`. When successive layers are placed so that spheres of each layer are directly over a layer that is one layer away we get a stacking that we label `ABABABA....` and the result is a hexagonal closest packed structure. Thus in hexagonal closest -packing there are two alternating hexagonal layers `A` and `B` offset from each other so that the spheres in one layer sit in the small triangular depressions of neighbouring layers. |
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| 1398. |
Which of the following represents the closet packed arrangement of uniform solid spheres:A. simple cubic unit cellB. body centered cubic unit cellC. face centered cubic unit cellD. hcp unit cell |
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Answer» Correct Answer - C::D |
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| 1399. |
Which of the following statement is correct for a closet packed structure? (i) Each tetrahedral void is surrounded by 4 spheres and each sphere is surrounded by 8 tetrahedral voids. (ii) Each octahedral voids is surrounded by 6 spherus and each sphere is surrounded by 6 octahedral voids. (iii) The number of tetrahedral voids in a closest-packed arrangement is twice the number of spheres. (iv) The number of octahedral voids in a closest-packed arrangement is equal to the number of spheres.A. (i), (ii)B. (iii), (iv)C. (i), (iii)D. (i), (ii), (iii), (iv) |
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Answer» Correct Answer - 4 Tetrahedral voids are surrounded `4` spheres sitting at the corners of a regular tetrahedron and are present in `HCP`as well as `C CP`. for every sphere in the closest packed arrangement there is one octahedral void and two tetrahedral voids. The octahedral voids are larger than the tetradral voids. |
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| 1400. |
The arrangement of the first two layers, one above the other, in hep and ccp arrangement isA. Exactly same in both casesB. Partly same and partly differentC. Different from each otherD. Nothing definite |
| Answer» Correct Answer - 1 | |