a metallic wire is stretched so that its length increases by 2%. calculate the percentage change in resistance.
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Answer:
A wire us stretched increase to length by 5% consider to initial length is l and cross sectional area is A
∴ volume V=Al
Now length is increased by 5%
∴ New length =
100
105
=
20
21
l.
If new area is l considering that the volume of the wire is unchanged
V=A l=A;
20
21
l
⇒ A=
21
20
A
Now if resistanty of the wire is ρ then initial Resistance
R
1
=ρ
A
l
Final Resistance
R
2
=ρ
21
20
A
20
21
l
=
A
ρl
(
20
21
)
2
∴
R
1
R
2
=(
20
21
)
2
=
400
441
% change in resistance =(
R
1
R
2
−R
1
)×100%
=10.25%
Answered by
0
A wire us stretched increase to length by 5% consider to initial length is l and cross sectional area is A
∴ volume V=Al
Now length is increased by 5%
∴ New length =
100
105
=
20
21
l.
If new area is l considering that the volume of the wire is unchanged
V=A l=A;
20
21
l
⇒ A=
21
20
A
Now if resistanty of the wire is ρ then initial Resistance
R
1
=ρ
A
l
Final Resistance
R
2
=ρ
21
20
A
20
21
l
=
A
ρl
(
20
21
)
2
∴
R
1
R
2
=(
20
21
)
2
=
400
441
% change in resistance =(
R
1
R
2
−R
1
)×100%
=10.25%
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