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IIT JEE Advanced 2006 Paper 1
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© examsnet.com
Question : 20
Total: 120
A ball moves over a fixed track as shown in the figure. From A to B, the ball rolls without slipping. Surface BC is frictionless.
K
A
,
K
B
and
K
C
are kinetic energies of the ball at A, B and C, respectively. Then
h
A
>
h
C
;
K
B
>
K
C
h
A
>
h
C
;
K
C
>
K
A
h
A
=
h
C
;
K
B
=
K
C
h
A
<
h
C
;
K
B
>
K
C
Validate
Solution:
Using conservation of energy, we can write the following:
E
A
=
m
g
h
A
+
K
A
(1)
E
B
=
K
B
(2)
E
C
=
m
g
h
C
+
K
C
(3)
Since
E
A
=
E
B
=
E
C
, we have the following:
K
B
>
K
A
K
B
>
K
C
From Eqs. (1) and (3), we get
E
A
−
E
C
= 0
That is,
m
g
(
h
A
−
h
C
)
+
[
K
A
−
K
C
]
= 0
h
A
−
h
C
=
K
C
−
K
A
m
g
From kinetic energy
K
B
, only translational part gets converted into potential energy at point C, we get
h
C
<
h
A
. Therefore,
K
C
>
K
A
© examsnet.com
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