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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 materialrho_(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^(@) Cassume that the resistivity of the new material is equal to the sum of the resistivity ofits component materials. |
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Answer» Solution :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 |
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