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IIT JEE Advanced 2006 Paper 1
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© examsnet.com
Question : 12
Total: 120
Consider a cylindrical element as shown in the figure. The current that flows through the element is I and resistivity of material of the cylinder is ρ. Choose the correct option out the following:
Power loss in first half is four times the power loss in second half.
Voltage drop in first half is twice of voltage drop in second half.
Current density in both halves is equal.
Electric field in both halves is equal.
Validate
Solution:
We have
R =
ρ
l
A
=
ρ
l
π
r
2
R
1
R
2
=
A
2
A
1
=
(
2
r
4
r
)
2
=
(
1
2
)
2
=
1
4
P
1
P
2
=
I
2
R
1
I
2
R
1
=
1
4
Now,
P
2
=
4
P
1
V
1
V
2
=
I
R
1
I
R
2
=
1
4
J
1
J
2
=
A
2
2
A
1
=
1
4
E
1
E
1
=
J
1
J
2
=
1
4
Therefore, the voltage drop in the first half is twice that of in the second half of the element.
© examsnet.com
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