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JEE Main Physics Class 12 Current Electricity Part 1 Questions
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© examsnet.com
Question : 21
Total: 100
A constant voltage is applied between two ends of a metallic wire. If the length is halved and the radius of the wire is doubled, the rate of heat developed in the wire will be:
[Online April 15, 2018]
Increased 8 times
Doubled
Halved
Unchanged
Validate
Solution:
Rate of heat i.e., Power developed in the wire
=
P
=
V
2
R
Resistance of the wire of length,
L
R
1
=
ρ
L
A
=
ρ
L
π
r
2
∴
Power,
P
1
=
V
2
R
1
Resistance of the wire when length is halved i.e.,
L
∕
2
R
2
=
ρ
L
2
π
(
2
r
)
2
=
ρ
L
π
8
r
2
=
R
1
8
∴
Power,
P
2
=
V
R
1
8
=
8
V
R
1
or,
P
2
=
8
P
1
i.e., power increased 8 times of previous or original wire.
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