A copper wire is stretched to make it 0.1% layer. The % increase in resistance will be?
Answers
Answer:
Length=l
Cross−sectionarea=A
Resistance=R
After increase, Length=l
′
Cross-sectional area=A
′
Resistance=R
′
l
′
=l+0.1%l=l+
100
0.1
l=1.001l
But, volume remains same.
V
′
=V⇒A
′
l
′
=Al
⇒A
′
(1001)L=Al
∴A
′
=
1.001
A
New Resistance(R
′
)=ρ
A
′
l
′
R
′
=ρ
(A/1.001)
(1.001l)
R
′
=(1.001)
2
⋅R=(1.002)R
Now, percentage change in resistance=
R
R
′
−R
×100
=
R
(1.002)R−R
×100%=0.002×100%=0.2%
Answer:
Assume initially,
Length=l
Cross−sectionarea=A
Resistance=R
After increase, Length=l
′
Cross-sectional area=A
′
Resistance=R
′
l
′
=l+0.1%l=l+
100
0.1
l=1.001l
But, volume remains same.
V
′
=V⇒A
′
l
′
=Al
⇒A
′
(1001)L=Al
∴A
′
=
1.001
A
New Resistance(R
′
)=ρ
A
′
l
′
R
′
=ρ
(A/1.001)
(1.001l)
R
′
=(1.001)
2
⋅R=(1.002)R
Now, percentage change in resistance=
R
R
′
−R
×100
=
R
(1.002)R−R
×100%=0.002×100%=0.2%
∴ Resistance will be increased by 0.2%.
Explanation:
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