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# If a copper wire is stretched to make it 0.1%longer. The percentage change in its resistance is

Athul Joseph
17 Points
3 years ago
Volume is constant during stretching.A*l=constantpi*r^2*l=constantl inversely proportional to r^2Using errors(Дl/l)*1000.-2*(Дr/r)*100Change in radius is equal to 0.2%Using errors again% change in resistance=4*(Дr/l)*100=4*0.2=0.8%
Khimraj
3007 Points
3 years ago
As resistance is directly proportional to the square of the length keeping its volume constant then by diffrentiating partially on both side we gey dR/R= 2*dL/L. But given dL/L=0.1 so persentage change in the resistance dR/R=2*0.1=0.2%
Athul Joseph
17 Points
3 years ago
For small change in length bu stretching(less than 8%)u can use the equationdR/R=4*dl/L.so the answer is 4*0.1=0.4.if u need derivation just ask.my previous solution is wrong.sorry.😀