What length of a copper wire of cross-sectional area `0.01 mm^2` will be needed to prepare a resistance of `1 kOmega`? Resistivity of copper `= 1.7 xx 10^(-8) Omega m`.

Explanation

`rho _cu = 1.7 xx (10^-8) w-m, `
` A = 0.01 mm^2 = 0.01 xx (10^-6) (m^2)`
` R = 1 kw = 10^3 w `
` So, R = rho l/ a `
` rArr (10^3) = (1.7 xx (10^-8) xx l / (10^-8))`
` l = (10^3 )/1.7 = 0.58 xx (10^3)m`
` = 0.6 km` .