Correct Answer - `10^(-4) ms^(-1)`
Here, density, `d = 8.9 xx 10^(3) kg m^(-3), M = 63.5 kg`,
`N = 6.02 xx 10^(26)` per kg atom , `I = 1.5 A`
`D = 1.2 mm = 1.2 xx 10^(-3) m`
Mass of `1m^(3)` of copper `= 8.9 xx 10^(3) kg`
Number of atoms per `m^(3)` of copper `= (N)/(M//d) = (Nd)/(M)`
As 1 atoms contributes one conduction electron, therefore number of conbination electron per `m^(3)` of copper is
`n = (Nd)/(M) = ((6.20 xx 10^(26)) xx (8.9 xx 10^(3)))/(63.5)`
`= 8.44 xx 10^(28)m^(-3)`
`upsilon_(d) = (I)/(n A e)`
`=(1.5)/((8.44xx10^(28))xx[(22)/(7)xx(1.2 xx10^(-3))^(2)/(4)]xx(1.6 xx10^(-19)))`
`= 0.98 xx 10^(-4) = 10^(-4) ms^(-1)`