Examsnet
Unconfined exams practice
Home
Aptitude
Algebra and Higher Mathematics
Arithmetic
Commercial Mathematics
Geometry and Mensuration
Number System and Numeracy
Problem Solving
Exams
Banking Entrance Exams
Defence Exams
Engineering Exams
Finance Entrance Exams
GATE Exams
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
State Eligibility Tests
UPSC Entrance Exams
CBSE
CBSE 10 Papers
CBSE Class 12 Previous Papers
NCERT Text Book Class 11 Solutions
NCERT Text Book Class 12 Solutions
ICSE Class 10 Papers
English
Competitive English
Certifications
Technical
Cloud Tech Certifications
Security Tech Certifications
Management
IT Infrastructure
More
About
Careers
Contact Us
Our Apps
Privacy
Test Index
MHT CET 2020 Physics Solved Paper
Show Para
Hide Para
Share question:
© examsnet.com
Question : 1
Total: 50
For the same cross-sectional area and for a given load, the ratio of depressions for the beam of a square cross-section and circular cross-section is
3
:
π
π
:
3
1
:
π
π
:
1
Validate
Solution:
δ
=
W
ℓ
3
3
Y
I
, where
W
=
load,
ℓ
=
length of beam and
I
is geometrical moment of inertia for rectangular beam,
I
=
b
d
3
12
where
b
=
breadth and
d
=
depth
For square beam
b
=
d
∴
I
1
=
b
4
12
For a beam ofcircular cross-section,
I
2
=
(
π
r
4
4
)
∴
δ
1
=
W
ℓ
3
×
12
3
Y
b
4
=
4
W
ℓ
3
Y
b
4
(for sq. cross-section)
and
δ
2
=
W
ℓ
3
3
Y
(
π
r
4
∕
4
)
=
4
W
ℓ
3
3
Y
(
π
r
4
)
(for circular cross-section)
Now
δ
1
δ
2
=
3
π
r
4
b
4
=
3
π
r
4
(
π
r
2
)
2
=
3
π
(
∵
b
2
=
π
r
2
i.e., they have same cross-sectional area)
© 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
Next Question
Examsnet
