Examsnet
Unconfined exams practice
Home
Exams
Banking Entrance Exams
CUET Exam Papers
Defence Exams
Engineering Exams
Finance Entrance Exams
GATE Exam Practice
Insurance Exams
International Exams
JEE Exams
LAW Entrance Exams
MBA Entrance Exams
MCA Entrance Exams
Medical Entrance Exams
Other Entrance Exams
Police Exams
Public Service Commission (PSC)
RRB Entrance Exams
SSC Exams
State Govt Exams
Subjectwise Practice
Teacher Exams
SET Exams(State Eligibility Test)
UPSC Entrance Exams
Aptitude
Algebra and Higher Mathematics
Arithmetic
Commercial Mathematics
Data Based Mathematics
Geometry and Mensuration
Number System and Numeracy
Problem Solving
Board Exams
Andhra
Bihar
CBSE
Gujarat
Haryana
ICSE
Jammu and Kashmir
Karnataka
Kerala
Madhya Pradesh
Maharashtra
Odisha
Tamil Nadu
Telangana
Uttar Pradesh
English
Competitive English
Certifications
Technical
Cloud Tech Certifications
Security Tech Certifications
Management
IT Infrastructure
More
About
Careers
Contact Us
Our Apps
Privacy
Test Index
KEAM 2019 Physics and Chemistry Paper
Show Para
Hide Para
Share question:
© examsnet.com
Question : 105
Total: 120
The edge length of a solid possessing cubic unit cell is
2
√
2
r
(structure I), based on hard sphere model, which upon subjecting to a phase transition, a new cubic structure (structure II) having an edge length of
4
√
3
r
is obtained, where r is the radiusof the hard sphere. Which of the following statements is true?
Density of the structure II is lower than structure I
Density of structure II is higher than structure I
The pore volume in structure I is 1.2 times higher than that of structure II
The pore volume of both the structures are equal
The octahedral voids in structure I is transformed into tetrahedral voids in structure II
Validate
Solution:
Density of a unit cell
∝
1
a
3
(
∵
z
=
1
)
for structure
I
,
d
1
∝
1
(
2
√
2
r
)
3
....(i)
for structure II
d
2
∝
1
(
4
r
√
3
)
3
.......(ii)
Dividing Eqs. (i) and (ii)
d
1
d
2
=
1
(
2
√
2
)
3
r
3
1
(
4
√
3
)
3
×
r
3
⇒
d
1
d
2
=
1
16
(
√
2
)
×
√
3
×
√
3
×
√
3
4
=
3
√
3
64
√
2
d
1
d
2
⇒
5.19
90.5
d
1
=
d
2
×
5.19
90.5
⇒
d
1
=
0.056
×
d
2
Final result concludes that density of structure I is higher than structure II.
Hence, statement I is correct.
© examsnet.com
Go to Question:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
Prev Question
Next Question