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Grade: 12
        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
3 years ago

Answers : (3)

utpal raj
244 Points
							
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 . 
hope this will clear your doubt , please aproove 
good luck
3 years ago
Kshitij Sharma
33 Points
							
2%. because the formula isR=\rho L/A.\RightarrowAR=\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\RightarrowLdA=-AdL.....(2)  .From(1) \frac{AdL-LdA}{LA}=\frac{dR}{R}​And from (2) \frac{dR}{R}=\frac{2dL}{L}​ = 2 X 1 % =2 %
3 years ago
arYa kumaR
39 Points
							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]
2 years ago
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