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Question

If a copper wire is stretched to make its cross sectional radius 0.1% thinner,then what is % increase in its resistence?

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Solution

Step 1, Given data

Radius becomes 0.1% thinner

Step 2, Finding the percentage increases in resistance

We know,

Resistance of the wire i given as

R=ρLA Where, ρ=Resistivity, L= length, A = Cross-sectional area.

We can say that due to the stretching of the wire volume of the wire will remain constant

We can write

R=ρLA=ρL2AL=ρL2V

In the above equation, ρ and V are constant

So,

ΔRR=2ΔLL

Or,%(ΔRR)=2%(ΔLL)=2×0.1=0.2%

Hence the percentage increase in resistance is 0.2 %



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