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JEE Advanced Model Paper 3 with solutions for online practice
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© examsnet.com
Question : 3
Total: 60
A ring of moment of inertia
10
−
2
k
g
m
2
about an axis passing through its centre and perpendicular to its plane is placed with its centre at origin. The current is passed through ring so that magnetic moment is
→
M
=
(
4
^
i
−
3
^
j
)
A -
m
2
. When magnetic field
→
B
=
(
3
^
i
+
4
^
j
)
is switched on at t=0. Then the angular acceleration of ring at t = 0 is 5 ×
10
n
ad/
s
2
then the value of n is
Your Answer:
Validate
Solution:
→
τ
=
→
M
×
→
B
= (
4
^
i
−
3
^
j
) × (
3
→
i
+
4
^
j
) = 25
^
k
(N - m)
Moment of inertial about an axis passing through centre of mass and parallel to
→
τ
.
I =
10
−
2
2
(Kg .
m
2
)
α =
τ
I
=
25
0.5
×
10
24
= 50000 rad /
s
e
c
2
© examsnet.com
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