when length of wire is increased by 10% then percentage increase in resistance is ?
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Answered by
1
If volume and density remains same, then resistance of wire:
R∝l
2
where l is length of the wire
∴R
′
=(1+
100
10
l)
2
=(
10
11
)
2
R
Hence,
R
R
′
−R
=
100
21
=21%
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2
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