1.

On applying an electric field of 15xx 10^(-6) vm^(-1)across a conductor , current density through it is 3.0 Am^(-2). The resistivity of the conductor is .......

Answer»

`45 XX 10^(-6) Omega m`
`5 xx 10^(-6) Omega `m
`0.5 xx 10^(-6) Omega m `
`2 xx 10^(5) Omega` m

Solution :`5 xx 10^(-6) Omega `m
`J = sigma ErArr (E)/(RHO)`
`THEREFORE rho= (E)/(J) = (15 xx 10^(-6))/(3) = 5 xx 10^(-6) Omega`m


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