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# A copper wire is stretched to increase its length by .1% . what is the percentage change in the resistance of the wire. Pleas give me full details

utpal raj
244 Points
4 years ago
hii rajasanu
r=pl/A  where p=resistivity of material , l=length of material, A = area of material
then you stretch length by 1% l1 = l+0.01l  =1.01l
find r1=pl1/A  and r2= pl/A  and r1-r2 will be the change in resistance .
good luck
Kshitij Sharma
33 Points
4 years ago
2%. because the formula is$R=\rho L/A$.$\Rightarrow$$AR=\rho L$$\Rightarrow$$\log (AR)=\log (\rho L)$$\Rightarrow$$\log A+\log R=\log \rho +\log L$
On partial differentiation, we get $\frac{dA}{A}+\frac{dR}{R}=\frac{d\rho }{\rho }+\frac{dL}{L}$$.....(1)$.since $\rho$​ is const,$d\rho$​ = 0 and d(volume)=d(LA)=0$\Rightarrow$​LdA+AdL=0$\Rightarrow$$LdA=-AdL.....(2)$  .From(1) $\frac{AdL-LdA}{LA}=\frac{dR}{R}$​And from (2) $\frac{dR}{R}=\frac{2dL}{L}$​ = 2 X 1 % =2 %
arYa kumaR
39 Points
4 years ago
let Rinitial =R1 = R and Rfinal =R2
length= L1 =L and L2 = 1001L/1000
since wire is same, V1 =V2
A1.L1=A2.L2
L1/L2=A2/A1
Now R1/R2= ∫L1/A1 × ∫L2/A2
R2 = L1.L1 /L2.L2 × R
= 1002001/1000000 R
Change in R = R2-R1
= 2001/1000000 R
Change % = 2001/1000000 R × 1/R × 100
= 0.2001% = 0.2%[APPROX]