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JEE Main 16 March 2021 Shift 1 Solved Paper
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© examsnet.com
Question : 11
Total: 90
Four equal masses,
m
each are placed at the corners of a square of length
(
I
)
as shown in the figure
[16 Mar 2021 Shift 1]
The moment of inertia of the system about an axis passing through
A
and parallel to
D
B
would be
1
m
l
l
2
2
m
l
2
3
m
l
2
√
3
m
l
2
Validate
Solution:
The given situation can be represented as follows
In the above figure, XX′ be the axis which passes through A and is parallel to DB.
∴Moment of inertia of the system
=
Mass
×
(Perpendicular distance from axis
)
2
I
=
m
(
A
C
)
2
+
m
(
E
D
)
2
+
m
(
F
B
)
2
=
m
(
0
)
2
+
m
(
I
√
2
)
2
+
m
(
‌
1
√
2
)
2
+
m
(
‌
1
√
2
)
2
=
3
m
l
2
© examsnet.com
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