A copper wire has diameter 0.5mm and resistivity of `1.6 times 10^-8 Omega.m` What will be the length of this wire to make its resistances `10 Omega`? How much does the resistance change if the diameter is doubled?

Explanation

It is given that diameter of wire D=0.5mm `= 5 times 10^-4m` resistivity `p=1.6 times 10^-8 Omega m`. And resistance `R=10 Omega`
As `(pL)/A=(4pL)/(piD^2),hence I=(piD^2R)/(4p)`

`therefore` Length of wire `L=(22 times( 5 times10^-4)^2 times10)/(7 times 4 times 1.6 times 10^-8)=122.5m`
IF the diameter of wire is doubled then `D.=2D` and hence `A.=(piD^2)/4=pi/4(2D^2)=4 times(piD^2)/4=4A`
For a given length and given material resistance is inversely proportional to the cross section area of the wire i.e., ` R prop 1/A`
`therefore (R.)/R=A/(A.)or R.=( RA)/(A.)=R/4=10/4=2.5 Omega`