Two materials have the value of `alpha_(1) and alpha_(2)` as `6 xx 10^(-4) (""^(@) C )^(-1) and - 5 xx 10^(-4) (""^(@) C )^(-1)`
respectively. The resistivity of the first material `rho_(20) = 2 xx 20^(-8) Omega. ` A new material is made by combining the above two materials. the resistivity does not change with temperature . The resistivity `rho_(20) ` of the second material is .... Considering the reference temperature as `20^(@) C ` assume that the resistivity of the new material is equal to the sum of the resistivity of its component materials.

Explanation

Here the reference temperature is `20" "^(@)` C.
Resistivity of a material at temperature `theta` is ,
`rho_(theta ) = rho_(20) [ 1 + alpha (theta - 20)]`
`rho theta = rho 20 + rho_(20) alpha theta - rho_(20) infty 20 `
` therefore (d rho theta)/(d theta ) = rho_(20) alpha `
For first material `((d rho theta )/(d theta ) )_(1) = (rho_(20) )_(1) alpha_(1)`
For second material `((d rho theta )/(d theta ) )_(2) = (rho_(20) )_(2) alpha_(2)`
The resistivity of the mixture `rho_(theta) = (rho theta )_(1) + (rho theta )_(2)` does not change with temperature . Therefore ,
`(d rho theta )/(d theta ) = ((d rho theta )/(d theta ))_(1) + ((d rho theta )/(d theta ))_(2) = 0 `
`therefore ((d rho theta )/(d theta ))_(1) = - ((d rho theta )/(d theta ))_(2) `
`therefore (rho_(20))_(1) alpha_(1) = - (rho_(20))_(2) alpha_(2)`
`therefore (rho_(20))_(2) = - ((rho_(20))_(1) alpha_(1))/(alpha_(2))`
= - ` ((2 xx 10^(-8) ) ( 6 xx 10^(-4)))/(- 5 xx 10^(-4))`
`therefore (rho_(20)_(2) = 2.4 xx 10^(-8) Omega` m