A uniform copper wire of mass `2.23 xx10^(-3)` kg carries a current of 1 A when 1.7 V is applied across it. Calculate the length and the area of cross section, if the wire is uniformaly stretched to double its length, calculate the new resistance. Density of copper is `8.92xx10^3 kgm^(-3)` and resistivity is `1.7xx10^(-8) Omega m.`

Explanation

As `m = volume xx density = (LxxS)xxd,`
`("as volume" = LxxS)`
`LxxS = (m)/(d) = (2.23xx10^(-3))/(8.92xx10^(3))=(1)/(4)xx10^(-6)` …(i)
And as V = IR, i.e.
`R =(V)/(I) =(1.7)/(1) = 1.7 Omega ….(ii)`
But by definition, …..(ii)
`R = rho(L//S)`
or `(L)/(S) = (R)/(rho) = (1.7)/(1.7xx10)^(-8) = 10^8 .....(iii)`
Solving Eqs. (i) and (iii) for L and S, we get L =5 m and
`S =5 xx10^(-8) m^2.` When the wire is stretched uniformly to double
its length, the volume will ramain unchanged, i.e.,
`SL = S'(2L) or S' = S//2`
Hence, the new resistance will be
`R' = rho(2L)/((S//2)) = 4rho (L)/(S) = 4R = 4.1xxx1.7 = 6.8 Omega`