The resistance of a wire is `20Omega`. What will be new resistance, if it is stretched uniformly 8 times its original length?

Explanation

`R_1 =20 Omega , R_2=?`
Let the original length `(l_1)` be l.
The new length `L_2 =8L_1(i.e)l_2=8l`
The original resistance.
`R_1 =rho(l_1)/(A_1)`
The new resistance `R_2=rho(l_2)/(A_2)=(rho(8l))/(A_2)`
Though the wire is stretched its volume is unchanged
Initial volume =Final volume
`A_1l_1=A_2I_2,A_1l=A_28l`
`(A_1)/(A_2)=(8l)/l=8`
By dividing equation `R_2` by equation `R_1` we get
`(R_2)/(R_1) =(rho(8l))/(A_2)xx(A_1)/(rhol) rArr (R_2)/(R_1)=(A_1)/(A_2)xx8`
Substituting the value of `(A_1)/(A_2)` we get
`(R_2)/(R_1)=8xx 8 =64 rArr R_2=64 xx20 =1280 Omega`
Hence, stretching the length of the wire has increased its resistance.