A copper wire has diameter 0.5 mm and resistivity of `1.6 xx 10^(-6) Omega` m. What will be the length of this wire to make its resistance `10Omega`? How much does the resistance change if the diameter is doubled?

Explanation

We are given,
the diameter of the wire
`d =0.5 mm =0.5 xx 10^(-3)m = 5 xx 10^(-4) m`
Resistivity of copper `rho =1.6 xx 10^(-8) Omega m`
Required resistance R = `10 Omega`
Length l =?
`As R = (rho l)/A`,
`l = (RA)/rho`
`=(R(pir^2))/rho ( :. A = pi r^2)`
`= (R (pid^2))/rho ( :. r = d/2)`
=`(pi R d^2)/(4rho)`
`=(3.14 xx 10 xx (5 xx 10^(-4))^2)/(4 xx 1.6 xx 10^(-8))`
`= (31.4 xx 25 xx 10^(-8))/(6.4 xx 10^(-8)) = 785/6.4 = 122.7 m`
Since resistance `R = (rho l)/A`
`= (rho l)/(pi r^2) =(rho l)/(pi(d/2)^2) =(4rho l)/(pid^2)`
`R prop 1/d^2` (If thereis no change in `rho and l`.)
So. when diameter d is doubled. then resistance R becomes one-fourth of its original value.