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IIT JEE Advanced 2009 Paper 2
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© examsnet.com
Question : 7
Total: 57
Two metallic rings A and B, identical in shape and size but having different resistivities
ρ
A
and
ρ
B
, are kept on top of two identical solenoids as shown in the figure below. When current I is switched on in both the solenoids in identical manner, the rings A and B jump to heights
h
A
and
h
B
, respectively, with
h
A
>
h
B
. The possible relation(s) between their resistivities and their masses m m A B and is(are)
ρ
A
>
ρ
B
and
m
A
=
m
B
ρ
A
<
ρ
B
and
m
A
=
m
B
ρ
A
>
ρ
B
and
m
A
>
m
B
ρ
A
<
ρ
B
and
m
A
<
m
B
Validate
Solution:
Induced emf is same in both the rings:
I =
e
R
=
ρ
A
ρ
l
I α
1
ρ
⇒ q α
1
ρ
(1)
Impulse is
J = ∫ Bil dt = mv = Bl∫ I dl = mv
That is, J = Blq = mv ⇒
v
(
q
m
)
(2)
and
v
2
∝ h (3)
From Eqs. (1), (2) and (3), we get
mv α
1
ρ
m
√
h
α
1
ρ
mρ α
1
√
h
Since
h
A
>
h
B
and for
m
A
=
m
B
,
ρ
√
h
= Constant. Therefore,
ρ
A
<
ρ
B
. Also if
m
A
<
m
B
and
ρ
A
<
ρ
B
,
m
A
ρ
A
<
m
B
ρ
B
⇒
h
A
>
h
B
(already given)
© examsnet.com
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