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JEE Main Physics Class 11 System of Particles and Rotational Motion Part 1 Questions
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© examsnet.com
Question : 26
Total: 100
Two identical spherical balls of mass
M
and radius
R
each are stuck on two ends of a rod of length
2
R
and mass
M
(see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is:
[10 Jan. 2019 II]
137
15
M
R
2
17
15
M
R
2
209
15
M
R
2
152
15
M
R
2
Validate
Solution:
For Ball
using parallel axes theorem, for ball moment of inertaia,
I
ball
=
2
5
M
R
2
+
M
(
2
R
)
2
=
22
5
M
R
2
For two balls
I
balls
=
2
×
22
5
M
R
2
=
and,
I
rod
=
M
(
2
R
)
2
12
=
M
R
2
3
I
system
=
I
balls
+
I
rod
=
44
5
M
R
2
+
M
R
2
3
=
137
15
M
R
2
© examsnet.com
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