A cube of side `0.05 m` of a material of conductor is drawn into a wire of thickness `0.20 mm`. If the specific resistance of the wire is `10^(-6)Omegam` then calculate the resistance of the wire.

Explanation

Since volume is conserved,
`a^(3)=AL=((pid^(2))/4)L,`
But, `R.=(pL)/A` where `L=(a^(3))/A`
`thereforeR.=(p/A)((a^(3))/A)=(a^(3)p)/(A^(2))`
where, `A=(pid^(2))/4` and `d=20xx10^(-3)m`
i.e., `R.=((5xx10^(-2))^(3)xx(10^(-6)))/(((3.142xx20xx10^(-5))/4)^(2))`
i.e., `R.=(125xx10^(-12)xx16)/(3.949xx10^(+3)xx10^(-10))` , i.e., `R.=0.506xx10^(-2)Omega`
or `R.=5.06mOmega`