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 uniformly 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

Let L be the length and A the cross sectional area of the wire
and d be the desity of the meterial of wire.
Masss M = Volume xx density = Ald
or `AL = (M)/(d) = (2.23xx10^(-3))/(8.92xx10^(3)) = (1)/(4)xx10^(-6)m^3`
Resistance of wire,
`R = (V)/(l) = (1.7)/(1) = 1.7 Omega`
`R = (rhoL)/(A) or (L)/(A) = (R )/(rho) = (1.7)/(1.7xx10^(-8)) = 10^8m^(-1) (ii)`
Multiplying (i) and (ii) we get
`L^(2) = (1)/(4)xx10^(-6)xx10^8 = (10^(2))/(4) or L = 5m`
From (i) `A = ((1)/(4)xx10^(-6))/(L)=((1)/(4)xx10^(-6))/(5) = (10^(-6))/(20) = 5xx10^(-8)m^(2)`
When wire is stretched to length L, its cross- sectional area
decreases. So `R prop L^(2)`. If R' is new resistacne of wire, then
`(R')/(R ) = ((L')/(L))^(2)`
or `R' = ((L')/(L))^(2)f R = (2)^(2)xx1.7 = 6.8Omega`