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Share/Save/Bookmark Login/ Register to Bookmark Topic : "Solid Doubt (s) Again" Started by Tush

Tush

#1 Posted 6:44pm 29-12-09  

Solid Doubt (s) Again

Ques 1) A body centred cubic lattice is made up of hollow spheres of B. Spheres of Solid A are present in hollow spheres of B. Radius of A is half of radius of B. What is the ratio of the total volume of spheres of B unoccupied by A in a unit cell and volume of unit cell ?

Ques 2) Select the correct statement(s) * :-
(a) Cs C l  changes to Na C l structure on heating.
(b) Na C l changes to Cs C l  structure on applying pressure.
(c) Co -ordination no. dec on applying pressure
(d) Co -ordination no. inc on heating.

* [hide]For Ques 2,  one / more answers is / are correct  [/hide]
Science is always wrong. It never solves a problem without creating ten more.    

Tush

#2 Posted 6:49pm 29-12-09  

Re: Solid Doubt (s) Again

yeah but what's the reason ???
Science is always wrong. It never solves a problem without creating ten more.    

Asish

#3 Posted 6:56pm 29-12-09  

Re: Solid Doubt (s) Again

When we heat things, they tend to expand (or molecules tend to vibrate more). So they will try to spread out. This can occur if the coordination no. decreases.

Conversely, on application of pressure, they will try to become more compact which is possible on increasing coordination no.
IIT has walked to me ... off to IIT K    

Tush

#4 Posted 7:37pm 29-12-09  

Re: Solid Doubt (s) Again

hhhhmm kk thanx Asish [1]
Science is always wrong. It never solves a problem without creating ten more.    

Tush

#5 Posted 0:37pm 03-01-10  

Re: Solid Doubt (s) Again

Please someone  try 1[p]st[/p] question.[2]
Science is always wrong. It never solves a problem without creating ten more.  Edited on 0:37pm 03-01-10    

JOHNCENA IS BACK

#6 Posted 0:48pm 03-01-10  

Re: Solid Doubt (s) Again

0.0154647393532935472636379137634458*2
coming soon!  Edited on 1:12pm 03-01-10    

Tush

#7 Posted 0:56pm 03-01-10  

Re: Solid Doubt (s) Again

Options given  :-
(a) 0.04    (b) 0.094   (c) 0.09158    (d) none of these


& Answer is given as[hide] (d)[/hide]
Science is always wrong. It never solves a problem without creating ten more.    

Ankur

#8 Posted 1:18pm 03-01-10  

Re: Solid Doubt (s) Again

Well solid state is really easy... I love that chapter!

In Qsn.1, first draw the structure of BCC wid spheres B. Radius of B = r, Side of cube = a.

Then, relating the length of diagonal and radius of B spheres, we get:

a = [frac]4√3[/]3[/frac]r


Total volume of A spheres is given by:

2 x  [frac]4∩ r[p]3[/p][/]3  8[/frac]

Hence remaining volume is 2 x [frac]4∩ 7 r[p]3[/p][/]3  8[/frac]

(Since total volume of outer spheres - total volume of inner spheres = volume unoccupied, i.e 1 - [frac]1[/]8[/frac] = [frac]7[/]8[/frac] )

Dividing this by volume of cube, a[p]3[/p]

We get:

2 x  [frac]4∩ 7 r[p]3[/p][/]3  8[/frac]
---------------     %
([frac]4√3[/]3[/frac]r )[p]3[/p]


Divide the two, cancel out common terms and you get answer as [b]0.5838642%[/b]

Just re-check my calculations and cancellations, cuz i think there might be some calculation error. Steps, otherwise, are perfect.

Hence answer according to my calculations should be [b][u]None of these.[/u][/b]
Careers, like rockets, don't always take off on schedule. The key is to keep working the engines.  Edited on 11:28pm 03-01-10    

OFFLINEEEEE!!

#9 Posted 1:23pm 03-01-10  

Re: Solid Doubt (s) Again

lol....jai ho wat outstanding answer :P

0.0154647393532935472636379137634458*2

simply 0.031!!!!!!!!!ha!!!!!ha!!!!!!!!
A.K.    

Avishek

#10 Posted 9:02pm 05-01-10  

Re: Solid Doubt (s) Again

Ankur can u xplain why have u multiplied the term wid 2 durin 'remaining volume calculation'?????
The spheres with radius A are within the one with radius B....the number of sphere wid radius B is 1.....other wise if there would 2 spheres wid radius B [b](one in the centre n half on each side)[/b] then the total nos of sheres with radius A would have been 4...
[i]Correct me if i am wrong...[/i]
Be brave, u wil always do well....the moment u r afraid u r nobody    

Ankur

#11 Posted 10:19pm 05-01-10  

Re: Solid Doubt (s) Again

Hey Avishek, let me give u a simple picture of what is my thought on this.

I considered one sphere of radius B, from which i remove the volume of sphere A, ok? So i get the volume of the remaining portion of the sphere of radius B.

[b]Since the qsn says its a BCC lattice, there are two spheres of radius B in the unit cell.
Since there are two such spheres, so we have to subtract the volume of sphere A from each of these 2 spheres. Hence i took 2 common from the 'remaining volume calculation'
[/b]

nd if u wanna know why,
its cuz, writing  1 - [frac]1[/]8[/frac] is better than writing 2 - [frac]2[/]16[/frac] aint it?

hope u got my point right... i guess my explanation must have been a bit confusing for ya.... in tht case srry fr the inconvenience [6]
Careers, like rockets, don't always take off on schedule. The key is to keep working the engines.    

Avishek

#12 Posted 11:19pm 07-01-10  

re:

yes i get it but i think there is som calculation mistake for which i was havng a prob [i]LET ME EXPLAIN IT WID SOLUTION[/i]



a=(4/√3)r

volume of sphere wid radius B = [frac]4 ∩ r[p]3[/p][/]3[/frac]

now A = [frac]1[/]2[/frac] B

therefore, there are two spheres wid radius a in a single one wid B

volume of a sphere wid radius A = [frac]4 ∩ ([frac]r[/]2[/frac])[p]3[/p][/]3[/frac]

which is  equal to [frac]4 ∩ r[p]3[/p][/]3 . 8[/frac]

now if we tak 1 sphere of radiud B it has 2 wid A

so, total volume of A's = [frac]2 4 ∩ r[p]3[/p] [/]3 . 8[/frac]

now the remaining volume of the one wid B = [frac]4 ∩ r[p]3[/p] [/]3[/frac] - [frac]2 4 ∩ r[p]3[/p] [/]3 . 8[/frac]

                                       = [frac]4 6 ∩ r[p]3[/p] [/]3 . 8[/frac]

now u can devide it wid 2 as there are 2 sphere wid radius B in a bcc structure.....
and the option is ofcourse 'd'

and the answer becomes ' .255 '

Be brave, u wil always do well....the moment u r afraid u r nobody    
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