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
JEE Main Physics Class 11 System of Particles and Rotational Motion Part 2 Questions
Show Para
Hide Para
Share question:
© examsnet.com
Question : 36
Total: 100
Consider a badminton racket with length scales as shown in the figure.
If the mass of the linear and circular portions of the badminton racket are same (M) and the mass of the threads are negligible, the moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at,
r
2
distance from the end A of the
handle will be ........
M
r
2
.
[26 Aug 2021 Shift 1]
Your Answer:
Validate
Solution:
The given situation is shown in the following figure
The moment of inertia of the racket about an axis perpendicular to the handle and in the plane of ring at,
r
2
distance from end A of the handle is given by
I
=
I
due
to
handle
+
I
due
to
disc
...(i)
I
due
to
handle
=
1
12
M
(
6
r
)
2
+
M
(
5
r
2
)
2
=
1
12
M
36
r
2
+
25
M
r
2
4
=
3
M
r
2
+
25
4
M
r
2
=
M
r
2
(
37
4
)
=
37
4
M
r
2
I
due
to
disc
=
M
r
2
2
+
M
(
13
r
2
)
2
=
M
r
2
2
+
169
M
r
2
4
=
171
Mr
2
4
∴ From Eq. (i), we get
I
=
I
dueto
handle
+
I
due
to
disc
=
37
4
M
r
2
+
171
M
r
2
4
=
208
4
M
r
2
=
52
Mr
2
© 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
Prev Question
Next Question
Examsnet