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IIT JEE Advanced 2007 Paper 2
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© examsnet.com
Question : 2
Total: 66
A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of
3
v
2
/4 g with respect to the initial position. The object is
ring
solid sphere
hollow sphere
disc
Validate
Solution:
From conservation of energy, we get
1
2
m
v
2
+
1
2
I
C
M
ω
2
= mgh
1
2
m
v
2
(
1
+
k
2
r
2
)
=
m
g
(
3
v
2
4
g
)
where k is the radius of gyration.
Now,
1
+
k
2
r
2
=
3
2
⇒
k
2
r
2
=
1
2
Therefore, the object is a disc.
© examsnet.com
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