If resistance `R_1` in resistance box is `300 Omega` then the balanced length is found to be `75.0 cm` from end `A`. The diameter of unknown wire is 1 mm and length of the unknown wire is `31.4 cm`. Find the specific resistance of the unknown wire

Explanation

`R/X=1/(100-l)`
`rarr X=((100-l)/l)R`
`=((100-75)/75) (300)=100Omega`
Now, `X=(rhol)/A=(rhol)/((pid^2//4))`
`rho=(pid^2X)/(4l)`
`=((22//7)(10^-3)^2(100))/((4)(0.314))`
`=2.5xx10^-4Omega-m`